In this article, we shall study to solve problems of probability based on the concept of the binomial distribution.

Example – 01:

An unbiased coin is tossed 5 times. Find the probability of getting a) three heads b) at least 4 heads

Solution:

In this case number of trials = n = 5, Probability of getting head (success) = 1/2

∴ p = 1/2 and q = 1 – p = 1 – 1/2 = 1/2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting exactly 3 heads (X = 3):

∴ P(X = 3) = ⁵C₃ (1/2)³ (1/2)^{5 – 3}

∴ P(X = 3) = ⁵C₃ (1/2)³ (1/2)²

∴ P(X = 3) = 10 x (1/2)⁵ = 10 x (1/32) = 5/16 = 0.3125

The probability of getting at least 4 heads (X ≥ 4):

∴ P(X ≥ 4) = P(X = 4) + P(X = 5)

∴ P(X ≥ 4) = ⁵C₄ (1/2)⁴ (1/2)^{5 – 4} + ⁵C₅ (1/2)⁵ (1/2)^{5 – 5}

∴ P(X ≥ 4) = 5 x (1/2)⁴ (1/2)¹ + 1 x  (1/2)⁵ (1/2)⁰

∴ P(X ≥ 4) = 5 x (1/2)⁵ + 1 x  (1/2)⁵

∴ P(X ≥ 4) = 5 x (1/32) + 1 x  (1/32) = 6/32 = 3/16 = 0.1875

Ans: The probability of getting exactly 3 heads is 5/16 or 0.3125 and the probability of getting at least 4 heads is 3/16 or 0.1875

Example – 02:

An unbiased coin is tossed 8 times. Find the probability of getting head a) exactly 5 times, b) a larger number of times than the tail, and c) at least once.

Solution:

In this case number of trials = n = 8, Probability of getting head (success) = 1/2

∴ p = 1/2 and q = 1 – p = 1 – 1/2 = 1/2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting exactly 5 heads (X = 5):

∴ P(X= 5) = ⁸C₅ (1/2)⁵ (1/2)^{8 – 5}

∴ P(X = 5) =  56 x (1/2)⁵ (1/2)³

∴ P(X = 5) =  56 x (1/2)⁸ = 56 x (1/256) = 56/256 = 7/32 = 0.2188

The probability of getting more heads than tail (X ≥ 5):

∴ P(X ≥ 5) = P(X = 5) + P(X = 6)  + P(X = 7) + P(X = 8)

∴ P(X ≥ 5) =  ⁸C₅ (1/2)⁵ (1/2)^{8 – 5} +  ⁸C₆ (1/2)⁶ (1/2)^{8 – 6} +  ⁸C₇ (1/2)⁷ (1/2)^{8 – 7}  + ⁸C₈ (1/2)⁸ (1/2)^{8 – 8}

∴ P(X ≥ 5) =  56 x (1/2)⁵ (1/2)³ + 28 x (1/2)⁶ (1/2)² +  8 x (1/2)⁷ (1/2)¹  + 1 x  (1/2)⁸ (1/2)⁰

∴ P(X ≥ 5) =  56 x (1/2)⁸  + 28 x (1/2)⁸ +  8 x (1/2)⁸  + 1 x  (1/2)⁸

∴ P(X ≥ 5) = (56 + 28 + 8 + 1)x  (1/256) = 93/256 = 0.3633 

The probability of getting atleast one head (X ≥ 1):

∴ P(X ≥ 1) =  1 – P(X = 0)

∴ P(X ≥ 1) = 1 –   ⁸C₀ (1/2)⁰ (1/2)^{8 – 0}

∴ P(X ≥ 1) =  1 x (1/2)⁰ (1/2)⁸

∴ P(X ≥ 1) =  1 – 1/256 = 255/256 = 0.9961

Ans: The probability of getting exactly 5 heads is 7/32 or 0.2188

The probability of getting a head a larger number of times than the tail is 93/256 or 0.3633

The probability of getting atleast one head is 255/256 or 0.9961

Example – 03:

An unbiased coin is tossed 9 times. Find the probability of getting head a) exactly 5 times, b) in the first four tosses, and tails in the last five tosses.

Solution:

In this case number of trials = n = 8, Probability of getting head (success) = 1/2

∴ p = 1/2 and q = 1 – p = 1 – 1/2 = 1/2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting exactly 5 heads (X = 5):

∴ P(X= 5) = ⁹C₅ (1/2)⁵ (1/2)^{9 – 5}

∴ P(X = 5) =  126 x (1/2)⁵ (1/2)⁴

∴ P(X = 5) =  126 x (1/2)⁹ = 126 x (1/512) = 63/256 = 0.2461

The probability of gettingin head in first four tosses and tails in last five tosses :

∴ P(X) = (1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = 1/512 = 0.001953

Ans: The probability of getting exactly 5 heads is 63/256 or 0.2461 and the probability of getting head in the first four tosses and tails in the last five tosses is 1/512 or 0.001953

Example – 04:

If the chance that out of 10 telephone lines one of the line is busy at any instant is 0.2. What is the chance that 5 of the lines are busy?

Solution:

In this case number of lines = n = 10, Probability of getting line busy (success) = 0.2

∴ p = 0.2 and q = 1 – p = 1 – 0.2 = 0.8

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting 5 lines busy (X = 5):

∴ P(X = 5) = ¹⁰C₅ (0.2)⁵ (0.8)^{10 – 5}

∴ P(X = 5) = ¹⁰C₅ (0.2)⁵ (0.8)⁵

∴ P(X = 5) = 252 x  (0.2 x 0.8)⁵ = 252 x (0.16)⁵ = 0.0264

Example – 05:

Each of five questions on a multiple-choice examination has four choices, only one of which is correct. The student is attempting to guess the answers. The random variable X is the number of questions answer correctly. What is the probability that the student will get a) exactly three correct answers? b) atmost three correct answers? c) at least one correct answer.

Solution:

In this case number of trials = n = 5, Probability of getting correct answer (success) = 1/4

∴ p = 1/4 and q = 1 – p = 1 – 1/4 = 3/4

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting exactly 3 answers correct (X = 3):

∴ P(X = 3) = ⁵C₃ (1/4)³ (3/4)^{5 – 3}

∴ P(X = 3) = 10 x (1/4)³ (3/4)²

∴ P(X = 3) = 10 x  (1/64) (9/16) = 90/1024 = 45/512 = 0.0879

The probability of getting atmost 3 correct answers (X ≤ 3):

∴ P(X ≤ 3) = P(X = 0) + P(X = 1)  + P(X = 2) + P(X = 3)

∴ P(X ≤ 3) = ⁵C₀ (1/4)⁰ (3/4)^{5 – 0} + ⁵C₁ (1/4)¹ (3/4)^{5 – 1}+ ⁵C₂ (1/4)² (3/4)^{5 – 2} +  ⁵C₃ (1/4)³ (3/4)^{5 – 3}

∴ P(X ≤ 3) = 1 x 1 x  (3/4)⁵+ 5 x  (1/4)¹ (3/4)⁴  + 10 x (1/4)² (3/4)³ + 10 x (1/4)³ (3/4)²

∴ P(X ≤ 3) = (243/1024) + 5 x  (1/4) x (81/256) + 10 x (1/16) (27/64) + 10 x (1/64) (9/16)

∴ P(X ≤ 3) = (243/1024) + (405/1024) + (270/1024) + (90/1024)

∴ P(X ≤ 3) = 1008/2024 = 63/64 = 0.9844

The probability of getting atleast 1 correct answers (X ≥ 1):

∴ P(X ≥ 1) = 1 – P(X = 0)

∴ P(X ≥ 1) = 1 –  ⁵C₀ (1/4)⁰ (3/4)^{5 – 0 3}

∴ P(X ≥ 1) = 1 –  1 x 1 x  (3/4)⁵

∴ P(X ≥ 1) = 1-  (243/1024) = 781/1024 = 0.7627

Ans: The probability of getting exactly 3 answers correct is 45/512 or 0.0879, the probability of getting atmost 3 correct answers is 63/64 or 0.9844, the probability of getting atleast 1 correct answer is 781/1024 or 0.7627

Example – 06:

The probability of hitting a target in any shot is 0.2. If 10 shots are fired, find the probability that the target will be heat atleast twice

Solution:

In this case number of trials = n = 10, Probability of hitting target (success) = 0.2

∴ p = 0.2 and q = 1 – p = 1 – 0.2 = 0.8

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of hitting the target atleast twice (X ≥ 2):

∴ P(X ≥ 2) = 1 – { P(X = 0) + P(X = 1)}

∴ P(X ≥ 2) = 1 – { ¹⁰C₀ (0.2)⁰ (0.8)^{10 – 0} + ¹⁰C₁ (0.2)¹ (0.8)^{10 – 1}}

∴ P(X ≥ 2) = 1 – { 1 x  1 x  (0.8)¹⁰ + 10 x  (0.2) (0.8)⁹}

∴ P(X ≥ 2) = 1 – (0.8 + 2) (0.8)⁹ = 1 – (2.8) (0.8)⁹

∴ P(X ≥ 2) = 1 – 0.3758= 0.6242

Ans: The probability of hitting the target atleast twice is 0.6242

Example – 07:

The probability that a bomb will hit a target is 0.8. Find the probability that out of 10 bombs dropped, exactly two will miss the target.

Solution:

In this case number of trials = n = 10, Probability of hitting target (success) = 0.8

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

Exactly two miss the target implies 8 bombs hit the target

The probability exactly two bombs miss the target  (X = 2):

∴ P(X = 8) = ¹⁰C₈ (0.8)⁸ (0.2)^{10 – 8}

∴ P(X = 8) = 45 x (0.8)⁸ (0.2)²

∴ P(X = 8) = 0.3020

Ans: The probability exactly two bombs miss the target is 0.3020

Example – 08:

In a town, 80% of all the families own a television set. If 10 families are interviewed at random, find the probability that a) seven families own a television set b) atmost three families own a television set.

Solution:

In this case number of trials = n = 10, Probability of family having atelevision st is 80% = 80/100 = 0.8

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability that 7 families have television (X = 7):

∴ P(X = 7) = ¹⁰C₇ (0.8)⁷ (0.2)^{10 – 7}

∴ P(X = 7) = 120 x (0.8)⁷ (0.2)³ = 0.2013

The probability that atmost 3 families have television (X ≤ 3):

∴ P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

∴ P(X ≤ 3) = ¹⁰C₀ (0.8)⁰ (0.2)^{10 – 0} + ¹⁰C₁ (0.8)¹ (0.2)^{10 – 1} + ¹⁰C₂ (0.8)² (0.2)^{10 – 2} + ¹⁰C₃ (0.8)³ (0.2)^{10 – 3}

∴ P(X ≤ 3) = 1 x  1 x (0.2)¹⁰ + 10 x  (0.8) (0.2)⁹ + 45 x (0.8)² (0.2)⁸ + 120 x (0.8)³ (0.2)⁷

∴ P(X ≤ 3) = 0.0008644

Ans: The probability that 7 families have television is 0.2013 and the probability that atmost 3 families have television is 0.0008644

Example – 09:

The probability that a person who undergoes a kidney operation will recover is 0.7, Find the probability that of the six patients who undergo similar operations: a) none will recover, b) all will recover, c) half of them will recover and iv) aleast half will recover.

Solution:

In this case number of trials = n = 6, Probability of recovery after operation (success) = 0.7

∴ p = 0.7 and q = 1 – p = 1 – 0.7 = 0.3

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability that none will recover (X = 0):

∴ P(X = 0) = ⁶C₀ (0.7)⁰ (0.3)^{6 – 0}

∴∴ P(X = 0) = 1 x 1 x (0.3)⁶  = 0.000729

The probability that all will recover (X = 6):

∴ P(X = 6) = ⁶C₆ (0.7)⁶ (0.3)^{6 – 6}

∴∴ P(X = 6) = 1 x (0.7)⁶ x 1 = 0.1176

The probability that halff of them will recover (X = 3):

∴ P(X = 3) = ⁶C₃ (0.7)³ (0.3)^{6 – 3}

∴ P(X = 3) = 20 x (0.7)³ (0.3)³ = 0.1852

The probability that atleast half of them will recover (X ≥ 3):

∴ P(X ≥ 3) = P(X = 3) + P(X = 4) + P(x = 5) + P(X = 6)

∴ P(X ≥ 3) = ⁶C₃ (0.7)³ (0.3)^{6 – 3}  +  ⁶C₄ (0.7)⁴ (0.3)^{6 – 4}  +  ⁶C₅ (0.7)⁵ (0.3)^{6 – 5}  +  ⁶C₆ (0.7)⁶ (0.3)^{6 – 6}

∴ P(X ≥ 3) = 20 x (0.7)³ (0.3)³  + 10 x (0.7)⁴ (0.3)²  +  6 x (0.7)⁵ (0.3)¹  + 1 x (0.7)⁶ (0.3)⁰

∴ P(X ≥ 3) = 0.9294

Ans: The probability that none will recover is 0.000729. The probability that all will recover is 0.00086441176. The probability that half of them will recover is 0.1852. The probability that atleast half of them will recover is 0.9294

Example – 10:

Centres for disease control have determined that when a person is given a vaccine, the probability that the person will develop immunity to a virus is 0.8. If eight people are given the vaccine, find the probability that a) none will develop immunity, b) exactly one will develop immunity, and c) all will develop immunity

Solution:

In this case number of trials = n = 8, Probability taht person develops immunity (success) = 0.78

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability that none will develop immunity (X = 0):

∴ P(X = 0) = ⁸C₀ (0.8)⁰ (0.2)^{8 – 0}

∴ P(X = 0) = 1 x 1 x (0.2)⁸  = 0.00000256

The probability that exactly 4 will develop immunity (X = 4):

∴ P(X = 4) = ⁸C₄ (0.8)⁴ (0.2)^{8 – 4}

∴ P(X = 4) = 70 x (0.8)⁴ (0.2)⁴  = 0.04587

The probability that all will develop immunity (X = 8):

∴ P(X = 8) = ⁸C₈ (0.8)⁸ (0.2)^{8 – 8}

∴ P(X = 8) = 1 x (0.8)⁸ x 1 = 0.1678

Example – 11:

A machine has fourteen identical components that function independently. It will stop working if three or more components fail. If the probability that the component fails is 0.1. Find the probability that the machine will be working.

Solution:

In this case number of trials = n = 14, Probability that component fails (success) = 0.1

∴ p = 0.1 and q = 1 – p = 1 – 0.1 = 0.9

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

Machine will stop working if three or more components fail.

Hence machine will be working if less than three components fail

The probability that machine is working (X < 3):

∴ P(X < 3) = P(X = 0) + P(X = 1) + P(x = 2)

∴ P(X < 3) = ¹⁴C₀ (0.1)⁰ (0.9)^{14 – 0} + ¹⁴C₁ (0.1)¹ (0.9)^{14 – 1} + ¹⁴C₂ (0.1)² (0.9)^{14 – 2}

∴ P(X < 3) = 1x 1 x  (0.9)¹⁴ + 14 x  (0.1)¹ (0.9)¹³ + 91 x  (0.1)² (0.9)¹²

∴ P(X < 3) = (1 x  (0.9)² + 14 x  0.1 x (0.9) + 91 x  (0.1)² )(0.9)¹²

∴ P(X < 3) = (0.81 + 1.26 + 0.91 )(0.9)¹²

∴ P(X < 3) =0.8416

Example – 12:

The probability that a person picked at random will support a constitutional amendment requiring an annual balanced budget is 0.8. If nine individuals are interviewed and they respond independently. What is the probability that at least two-thirds of them will support the amendment?

Solution:

In this case number of trials = n = 9, Probability that support the ammendment (success) = 0.8

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

atleast two third of nine i.e. atleast 6 supports the ammendment

The probability that two third support ammendment (X ≥ 6):

∴ P(X ≥ 6) = P(X = 6) + P(X = 7) + P(x = 8) + P(x = 9)

∴ P(X ≥ 6) = ⁹C₆ (0.8)⁶ (0.2)^{9 – 6} + ⁹C₇ (0.8)⁷ (0.2)^{9 – 7} + ⁹C₈ (0.8)⁸ (0.2)^{9 – 8} + ⁹C₉ (0.8)⁹ (0.2)^{9 – 9}

∴ P(X ≥ 6) = 84 x (0.8)⁶ (0.2)³ + 36 x (0.8)⁷ (0.2)² + 9 x (0.8)⁸ (0.2)¹ + 1x (0.8)⁹ (0.2)⁰

∴ P(X ≥ 6) = 84 x (0.8)⁶ (0.2)³ + 36 x (0.8)⁷ (0.2)² + 9 x (0.8)⁸ (0.2)¹ + 1x (0.8)⁹ x 1

∴ P(X ≥ 6) = 0.9143

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In this article, we shall study to find the probability of an event when data normally distributed is given.

Area Under Normal Curve (0 < x< z)

Example – 01:

A sample of 100 dry battery cells tested to find the length of life produced the following results. Mean = μ = 12 hours, standard deviation = σ = 3 hours. Assuming that the data are normally distributed, what percentage of battery cells are expressed to have the life a) more than 15 hours, b) less than 6 hours, and c) between 10 hours and 14 hours. Given:

Solution:

We have Mean = μ = 12 hours, standard deviation = σ = 3 hours, Total number of objects = N = 100

P(life of battery more than 15 hours)

The standardized value of x = 15 is z = (x – μ)/σ

= (15 – 12)/3 = 3/3 = 1

P(z >1) = area under the standard normal curve to the right of z = 1

P(z >1) = (Area to the right of z = 0) – (Area between z = 0 and z = 1)

P(z >1) = 0.5 – 0.3413 = 0.1587

P(life of battery less than 6 hours)

The standardized value of x = 6 is z = (x – μ)/σ

= (6 – 12)/3 = – 6/3 = – 2

P(z < – 2) = area under the standard normal curve to the left of z = -2

P(z < -2) = (Area to the left of z = 0) – (Area between z = 0 and z = -2)

P(z >1) = 0.5 – 0.4772 = 0.0228

P(life of battery between 10 hours and 14 hours)

The standardized value of x = 10 is z = (x – μ)/σ

= (10 – 12)/3 = – 2/3 = – 0.67

The standardized value of x = 14is z = (x – μ)/σ

= (14 – 12)/3 =  2/3 =  0.67

P(- 0.67 < z < 0.67) = area under the standard normal curve between

z = 0.67 and z = – 0.67

P(- 0.67 < z < 0.67) = (Area between z = 0 and z = – 0.67) + (Area between z = 0 and z = 0.67)

P(- 0.67 < z < 0.67) = 2 x (Area between z = 0 and z = – 0.67)

= 2 x 0.2486 = 0.4972

Ans: 15.87% batteries have life more than 15 hours.

2.28% batteries have life less than 6 hours.

49.72% batteries have life between 10 hours and 14 hours.

Example – 02:

In a certain examination, 500 students appeared. Means score is 68 and SD 8. Assuming that the data are normally distributed find the number of students scoring a) less than 50 and b) more than 60.

Solution:

We have Mean = μ = 68, standard deviation = σ = 8, Total number of students = N = 500

P(marks less than 50)

The standardized value of x = 50 is z = (x – μ)/σ

= (50 – 68)/8 = – 18/8 = – 2.25

P(z  < – 2.25) = area under the standard normal curve to the left of z = – 2.25

P(z < – 2.25) = (Area to the right of z = 0) – (Area between z = 0 and z = -2.25)

P(z < – 2.5) = 0.5 – 0.4878 = 0.0122

Number of students got less than 50 marks = N x P(z < – 2.25)

= 500 x 0.0122 = 6 (Appox.)

P(marks more than 60)

The standardized value of x = 50 is z = (x – μ)/σ

= (60 – 68)/8 = – 8/8 = – 1

P(z  > – 1) = area under the standard normal curve to the right of z = – 1

P(z > – 1) = (Area to the right of z = 0) + (Area between z = 0 and z = -1)

P(z > – 1) = 0.5 + 0.3413 = 0.8413

Number of students got more than 60 marks = N x P(z > – 1)

= 500 x 0.8413 = 421 (Appox.)

Ans: Number of students got less than 50 marks are 6

Number of students got more than 60 marks are 421

Example – 03:

Sacks of sugar-packed by an automatic loader having an average weight of 100 kg and with a standard deviation of 0.250 kg. Assuming that the data are normally distributed, find the chance of sack weighing less than 99.5 kg.

Solution:

We have Mean = μ = 100 kg, standard deviation = σ = 0.250 kg

P(weight less than 99.5 kg)

The standardized value of x = 99.5 is z = (x – μ)/σ

= (99.5 – 100)/0.250 = – 0.5/0.250 = – 2

P(z  < – 2) = area under the standard normal curve to the left of z = – 2

P(z < – 2) = (Area to the right of z = 0) – (Area between z = 0 and z = -2)

P(z < – 2) = 0.5 – 0.4772 = 0.0228

Ans: The chance of sack weighing less than 99.5 kg is 0.0228 or 2.28%

Example – 04:

In a test 0f 2000 electric bulbs, it was found that the life of a particular make was normally distributed with an average life of 2040 hours and a standard deviation of 60 hours. Estimate the number of bulbs likely to burn for a) between 1920 hours and 2160 hours and b) more than 2150 hours. Given A(2) = 0.4772, A(1.83) = 0.4664

Solution:

We have Mean = μ = 2040 hours, standard deviation = σ = 60 hours, Total number of bulbs = N = 2000

P(Working life between 1920 hours and 2160 hours)

The standardized value of x = 1920 is z = (x – μ)/σ

= (1920 – 2040)/60 = – 120/60 = – 2

The standardized value of x = 2160 is z = (x – μ)/σ

= (2160 – 2040)/60 = 120/60 = 2

P(- 2 < z < 2) = area under the standard normal curve between z = -2 and z = 2

P(- 2 < z < 2) = (Area between z = 0 and z = -2) + (Area between z = 0 and z = 2)

P(- 2 < z < 2)  = 2 x (Area between z = 0 and z = 2) = 2 x 0.4772 = 0.9544

Number of bulbs having life between 1920 hours and 2160 hours

= N x P(- 2 < z < 2)

= 2000 x 0.9544 =1909 (Appox.)

P(Working life morethan 2150 hours)

The standardized value of x = 2150 is z = (x – μ)/σ

= (2150 – 2040)/60= 110/60 = 1.83

P(z  > 1.83) = area under the standard normal curve to the right of z = 1.83

P(z > 1.83) = (Area to the right of z = 0) – (Area between z = 0 and z = 1.83)

P(z > 1.83) = 0.5 – 0.4664 = 0.0336

Number of bulbs having life more than 2150 hours

= N x P(z > 1.83)

= 2000 x 0. 0336 = 67 (Appox.)

Ans: The number of bulbs having life between 1920 hours and 2160 hours is 1909 and the number of bulbs having life more than 2150 hours is 67

Example – 05:

In a test of 2000 electric bulbs, it was found that the life of a particular make was normally distributed with an average life of 2040 hours and a standard deviation of 60 hours. Estimate the number of bulbs likely to burn for a) between 1920 hours and 2160 hours and b) more than 2150 hours. Given A(2) = 0.4772, A(1.83) = 0.4664

Solution:

We have Mean = μ = 2040 hours, standard deviation = σ = 60 hours, Total number of bulbs = N = 2000

P(Working life between 1920 hours and 2160 hours)

The standardized value of x = 1920 is z = (x – μ)/σ

= (1920 – 2040)/60 = – 120/60 = – 2

The standardized value of x = 2160 is z = (x – μ)/σ

= (2160 – 2040)/60 = 120/60 = 2

P(- 2 < z < 2) = area under the standard normal curve between z = -2 and z = 2

P(- 2 < z < 2) = (Area betweenz = 0 and z = -2) + (Area between z = 0 and z = 2)

P(- 2 < z < 2)  = 2 x (Area between z = 0 and z = 2) = 2 x 0.4772 = 0.9544

Number of bulbs having life between 1920 hours and 2160 hours

= N x P(- 2 < z < 2)

= 2000 x 0.9544 =1909 (Appox.)

P(Working life morethan 2150 hours)

The standardized value of x = 2150 is z = (x – μ)/σ

= (2150 – 2040)/60= 110/60 = 1.83

P(z  > 1.83) = area under the standard normal curve to the right of z = 1.83

P(z > 1.83) = (Area to the right of z = 0) – (Area between z = 0 and z = 1.83)

P(z > 1.83) = 0.5 – 0.4664 = 0.0336

Number of bulbs having life more than 2150 hours

= N x P(z > 1.83)

= 2000 x 0. 0336 = 67 (Appox.)

Ans: The number of bulbs having life between 1920 hours and 2160 hours is 1909 and the number of bulbs having life more than 2150 hours is 67

Example – 06:

The scores of 1000 students have a mean 14 and standard deviation 2.5. Assuming the data to be normally distributed, find a) how many students secured marks between 12 and 15? and b) How many student score more than 18. Given A(0.8) 0.2882, A(0.4) 0.1554, A(1.6) = 0.4452.

Solution:

We have Mean = μ = 14, standard deviation = σ = 2.5, Total number of students = N = 1000

P(score between 12 and 15)

The standardized value of x = 12 is z = (x – μ)/σ

= (12 – 14)/2.5 = – 2/2.5 = – 0.8

The standardized value of x = 15 is z = (x – μ)/σ

= (15 – 14)/2.5 = 1/2.5 = 0.4

P(- 0.8 < z < 0.4) = area under the standard normal curve between z = – 0.8 and z = 0.4

P(- 0.8 < z < 0.4) = (Area betweenz = 0 and z = – 0.8) + (Area between z = 0 and z = 0.4)

P(- 0.8 < z < 0.4)  = 0.2881 + 0.1554 = 0.4435

Number of students who score between 12 and 14 

= N x P(- 0.8 < z < 0.4)

= 1000 x 0.4435 = 444 (Appox.)

P(Score more than 18)

The standardized value of x = 18 is z = (x – μ)/σ

= (18 – 14)/2.5= 4/2.5 = 1.6

P(z  > 1.6) = area under the standard normal curve to the right of z = 1.6

P(z > 1.6) = (Area to the right of z = 0) – (Area between z = 0 and z = 1.6)

P(z > 1.6) = 0.5 – 0.4452 = 0.0548

Number of students who score more than 18 

= N x P(z > 1.6)

= 1000 x 0. 0548 = 55 (Appox.)

Ans: The number of students who score between 12 and 14 is 444 and the number of students who score more than 18 is 55

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The post Probability: Normal Distribution 02 appeared first on The Fact Factor.

The normal distribution refers to a family of continuous probability distributions described by the normal equation.

on the domain x ∈ (- ∞, ∞)

where x is a normal random variable, μ is the mean, σ is the standard deviation,

Thus the normal distribution can be completely specified by two parameter mean (μ) and standard deviation (σ) and is represented as N(μ, σ).

Mathematicians called this distribution a normal distribution, a physicist called it a Gaussian distribution, and scientists called it a bell curve due to its bell-like shape.

The normal distribution with mean μ = 0 and standard deviation, σ = 1 is called the standard normal distribution. It is denoted by N(0, 1).

Characteristics of a Normal Distribution

• The normal curve is symmetrical about the mean μ. It is perfectly symmetrical around its center. That is, the right side of the center is a mirror image of the left side.
• The mean is at the middle and divides the area into halves. The center of a normal distribution is located at its peak, and 50% of the data lies above the mean, while 50% lies below. It means that the mean, median, and mode are all equal in a normal distribution.
• There is also only one mode, or peak, in a normal distribution.
• Normal distributions are continuous and have tails that are asymptotic.
• The total area under the curve is equal to 1;
• It is completely determined by its mean and standard deviation (SD) σ (or variance σ²)
• Approximately 68 % of the data lies within 1 SD of the mean. Approximately 95 % of the data lies within 2 SD of the mean.  Approximately 99.7 % of the data lies within 3 SD of the mean.

The Z – score:

The number of standard deviations from the mean is called the standard score or z – score.

An arbitrary normal distribution can be converted to a standard normal distribution by changing variables to z.

Empirical Rules for z – Scores:

• Approximately 68 % of the data lies within 1 SD of the mean. 
• Approximately 95 % of the data lies within 2 SD of the mean. 
• Approximately 99.7 % of the data lies within 3 SD of the mean.

Importance of z – Score:

Z-Scores tell us whether a particular score is equal to the mean, below the mean or above the mean of a bunch of scores. They can also tell us how far a particular score is away from the mean.

We can use Z-scores to standardize scores from different groups of data. Then we can compare raw scores from different groups of data.

Example – 01:

95 % of students at the college are between 1.1 m and 1.7 m tall. Find mean and the standard deviation assuming a normal distribution.

Solution:

Mean = μ = average of the two values given = (1.1 + 1.7)/2 = 2.8/2 = 1.4 m

From empirical rule states that approximately 95 % of the data lies within 2 SD of the mean on either side.

Therefore the total deviation is 4 S.D.

4 S.D. = 1.7 – 1.1

4 S.D. = 0.6

1 S.D. = σ = 0.15 m

Ans: Mean = 1.4 m and the standard deviation is 0.15 m

Example – 02:

95 % of students in a class of 100 weigh between 62 kg and 90 kg. Find mean and standard deviation assuming a normal distribution.

Solution:

Mean = μ = average of the two values given = (62 + 90)/2 = 152/2 = 76 kg

From empirical rule states that approximately 95 % of the data lies within 2 SD of the mean on either side.

Therefore the total deviation is 4 S.D.

4 S.D. = 90 – 62

4 S.D. = 28

1 S.D. = σ = 7 kg

Ans: Mean = 76 kg and standard deviation is 7 kg

Example – 03:

68 % of marks of students in a certain test are between 51 and 64. Find mean and standard deviation assuming a normal distribution.

Solution:

Mean = μ = average of the two values given = (51 + 64)/2 = 115/2 = 57.5 kg

From empirical rule states that approximately 68 % of the data lies within 1 SD of the mean on either side.

Therefore the total deviation is 2 S.D.

2 S.D. = 64 – 51

2 S.D. = 13

1 S.D. = σ = 6.5 kg

Ans: Mean marks = 57.5 and standard deviation in marks is 6.5

Example – 04:

99.7 % of electrical components produced by a machine have lengths between 1.176 cm and 1.224 cm. Find mean and standard deviation assuming a normal distribution.

Solution:

Mean = μ = average of the two values given = (1.176 + 1.224)/2 = 2.4/2 = 1. 2 cm

From empirical rule states that approximately 99.7 % of the data lies within 3 SD of the mean on either side.

Therefore the total deviation is 6 S.D.

6 S.D. = 1.224 – 1.176

6 S.D. = 0.048

1 S.D. = σ = 0.008 cm

Ans: Mean length = 1.2 cm and standard deviation in length is 0.008 cm

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The post Probability: Normal Distribution appeared first on The Fact Factor.

In this article, we shall study to check whether the given function is a probability mass function or not.

Example – 01: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, 1 ≤ x ≤ 4) = P(1) + P(2) + P(3) + P(4)

∴   ∑ P(X = x, 1 ≤ x ≤ 4) = 0.1 + 0.2 + ).3 + 0.4 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.

Hence given function is a p.m.f.

Example – 02: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, 0 ≤ x ≤ 3) = P(0) + P(1) + P(2) + P(3)

∴   ∑ P(X = x, 0 ≤ x ≤ 3) = 0.5 + 0.2 + 0.18 + 0.12 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.

Hence given function is a p.m.f.

Example – 03: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that P(-1) = – 0.2 < 0 Hence the first condition is not satisfied.

Hence given function is not a p.m.f.

Example – 04: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, 0 ≤ x ≤ 1 = P(0) + P(1) + P(2)

∴   ∑ P(X = x, 0 ≤ x ≤ 2) = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1

Thus sum of all probabilities is not equal to1. Hence the second condition is not satisfied.

Hence given function is not a p.m.f.

Example – 05: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 2, 4, 6, 8) = P(2) + P(4) + P(6) + P(8)

∴   ∑ P(X = x, x = 2, 4, 6, 8) = 0.2 + 0.4 + 0.6 + 0.8 = 2 ≠ 1

Thus sum of all probabilities is not equal to 1. Hence the second condition is not satisfied.

Hence given function is not a p.m.f.

Example – 06: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that P(-1) = – 0.1 < 0 Hence the first condition is not satisfied.

Hence given function is not a p.m.f.

Example – 07: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

∴   P(X = 0) = 0²/5 = 0/5 = 0

∴   P(X = 1) = 1²/5 = 1/5

∴   P(X = 2) = 2²/5 = 4/5

The probability distribution is

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 0, 1, 2) = P(0) + P(1) + P(2)

∴   ∑ P(X = x, x = 0, 1, 2) = 0 + 1/5 + 4/5 = 5/5 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.

Hence given function is a p.m.f.

Example – 08: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

∴   P(X = 0) = (1- 1)/3 = 0/3 = 0

∴   P(X = 1) = (2 – 1)/3 = 1/3

∴   P(X = 2) = (3 – 1)/3 = 2/3

The probability distribution is

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 1, 2, 3) = P(1) + P(2) + P(3)

∴   ∑ P(X = x, x = 1, 2, 3) = 0 + 1/3 + 2/3 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.

Hence given function is a p.m.f.

Example – 09: 

The function is P( X = x) = (x – 5)/4, x = 5.5, 6.5, 7.5

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

∴   P(X = 5.5) = (5.5 – 5)/4 = 0.5/4 = 1/8

∴   P(X = 6.5) = (6.5 – 5)/4 = 1.5/4 = 3/8

∴   P(X = 7.5) = (7.5 – 5)/4 = 2.5/4 = 5/8

The probability distribution is

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 5.5, 6.5, 7.5) = P(5.5) + P(6.5) + P(7.5)

∴   ∑ P(X = x, x = 5.5, 6.5, 7.5) = 1/8 + 3/8 + 5/8 = 9/8 ≠ 1

Thus sum of all probabilities is not equal to 1. Hence the second condition is not satisfied.

Hence given function is not a p.m.f.

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In this article, we shall study to write probability mass function and to write probability distribution for the given event.

Example – 01:

If a coin is tossed two times and X denotes the number of tails. Find the probability distribution of X.

Solution:

If a coin is tossed two times. The sample space for the experiment is as follows

S = {HH, HT, TH, TT}

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails)

Probability mass function is

P( X = 0) = P(0) = 1/4

P( X = 1) = P(1) = 2/4 = 1/2

P( X = 2) = P(02) = 1/4
Hence, The probability distribution for the number of tails is as follows.

Example – 02:

If a coin is tossed three times and X denotes the number of tails. Find the probability mass function of X. Also write the probability distribution of X.
Solution:

If a coin is tossed three times. The sample space for the experiment is as follows

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = 1/8
P( X = 1) = P(1) = 3/8
P( X = 2) = P(2) = 3/8
P( X = 3) = P(3) = 1/8

The probability distribution for the number of tails is as follows.

Example – 03:

If a coin is tossed four times and X denotes the number of tails. Find the probability distribution of X.

Solution:

METHOD- I:

If a coin is tossed four times. The sample space for the experiment is as follows

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, TTHT, TTTH, TTTT}

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = 1/8
P( X = 1) = P(1) = 3/8
P( X = 2) = P(2) = 3/8
P( X = 3) = P(3) = 1/8
P( X = 4) = P(4) = 3/8

The probability distribution for the number of tails is as follows.

METHOD- II

If a coin is tossed four times. The sample space for the experiment is as follows

S = 2⁴ = 16

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = ⁴C₀ /16 = 1/16
P( X = 1) = P(1) = ⁴C₁ /16 = 4/16 = 1/4
P( X = 2) = P(2) = ⁴C₂ /16 = 6/16 = 3/8
P( X = 3) = P(3) = ⁴C₃ /16 = 4/16 = 1/4
P( X = 4) = P(4) = ⁴C₄ /16 = 1/16

The probability distribution for the number of tails is as follows.

Example – 04:

If a coin is tossed five times and X denotes the number of tails. Find the probability distribution of X.

Solution:

If a coin is tossed five times. The sample space for the experiment is as follows

S = 2⁵ = 32

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails) or 5 (five tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = ⁵C₀ /32 = 1/32
P( X = 1) = P(1) = ⁵C₁ /32 = 5/32
P( X = 2) = P(2) = ⁵C₂ /32 = 10/32 = 5/16
P( X = 3) = P(3) = ⁵C₃ /32 = 10/32 = 5/16
P( X = 4) = P(4) = ⁵C₄ /32 = 5/32
P( X = 5) = P(5) = ⁵C₅ /32 = 1/32

The probability distribution for the number of tails is as follows.

Example – 05:

If a coin is tossed six times and X denotes the number of tails. Find the probability distribution of X.

Solution:

If a coin is tossed six times. The sample space for the experiment is as follows

S = 2⁶ = 64

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails) or 5 (five tails) or 6 (six tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = ⁶C₀ /64 = 1/64
P( X = 1) = P(1) = ⁶C₁ /64 = 6/64 = 3/32
P( X = 2) = P(2) = ⁶C₂ /64 = 15/64
P( X = 3) = P(3) = ⁶C₃ /64 = 20/64 = 5/16
P( X = 4) = P(4) = ⁶C₄ /64 = 15/64
P( X = 5) = P(5) = ⁶C₅ /64 = 6/64 = 3/32
P( X = 6) = P(6) = ⁶C₆ /64 = 1/64

The probability distribution for the number of tails is as follows.

In the next article, we shall study problems in which we will be checking, whether the distribution given is probability mass function or not.

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