Normal distribution problem 1

Question (Normal distribution problems): x is normally distributed with mean 30 and variance of 16. Calculate the z score corresponding to x value of 40.

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Solution:

x is normally distributed. We convert the given value of x to standard normal z score using below equation.

Z = (x – µ)/ std. devn

Variance = 16

Standard deviation = 4

z = (40 – 30)/4 = 2.5

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Normal distribution problem 2

Question (Normal distribution problems): x is normally distributed with mean 30 and variance of 16. Calculate the probability of finding the value of x greater than 30.

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Solution:

x is normally distributed with mean 30.

Normal distribution is symmetrical around its mean.

Hence, the probability of finding x greater than 30 is 0.5

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Normal distribution problem 3

Question (Normal distribution problems): Height of children of a particular age group follows normal distribution with mean 60 inches and standard deviation of 6 inches. If eight children are selected at random from this age group, what is the probability that their average height is more than 64 inches?

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Solution:

Revise your concepts of Normal distribution here: Normal distribution curve and Normal distribution tables and areas

Height of children is normally distributed ~ N (60, 62)

Eight children are selected at random, so, sample size n = 8

Objective: to find probability on sample mean

Concept:

Sample mean from a normally distributed population with mean µ and standard deviation σ will follow a normal distribution with mean µ and standard deviation σ / √n where n is the sample size.

Sample mean is normally distributed ~ N (60, (6/√8)2)

P(X > 64) = P (Z > ((64 – mean)/standard deviation of the sample mean)

= P(Z > (64-60)/(6/√8)) = P(Z > 1.88) = 0.03

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Normal distribution problem 4

Question (Normal distribution problems): Suppose the IQ scores of all people who take the IQ test are normally distributed with a mean score 10 and standard deviation of 14.

1. What is the cutoff IQ score for top 2% of the people?
2. Suppose you take the test and if you get an IQ score of 124, what percentage of students fall below your score?

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Solution:

If you wish to revise the concepts of Normal distribution, refer to Normal distribution curve and Normal distribution tables and areas

1. We require cutoff score for top 2% of people. This indirectly translates to finding out the cutoff score where the are to the right of it is 2%.

Z value where the area to the right of it is 2% (0.02) is 2.05 (Normal distribution tables and areas) Z = 2.05 is the cutoff score. To convert this Z score to the raw IQ score, we use  Z = (x – µ)/ std. devn

2.05 = (x – 110)/14

x = 138.7

Hence, the cutoff score for top 2% IQ score is 138.7

2.

IQ Score (Raw) = 124

We first convert this raw score to Standard normal Z score using Z = (x – µ)/ std. devn

Z = (124 –  110)/14 = 1.0

Percentage of people who get score less than 124  = percentage of values less than 1.0 = Area to the left of 1.0 in z table.

Area to the left of 1.0 using z table is 0.8413 (Normal distribution tables and areas).

Hence, the percentage of people who have IQ scores less than 124 is 84.13%.

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Normal distribution problem 5

Question (Normal distribution problems) : Jack has recently started a firm that is primarily into manufacturing of spare parts for automobiles. Firm has automated the process of producing products so as to optimise the resource utilisation and expedite the process of delivering to customers. One of the key specifications for the product is that it must be 21 inches in length, the acceptable range being 20.5 inches to 21.5 inches.

Suppose the length of the goods produced follows a normal distribution with mean 21.6 inches and standard deviation of 1.1 inches

a) What is the probability that a randomly selected product conforms to the specifications of customer (In the range of 20.5 inches to 21.5 inches)

b) If a sample of 8 products is selected at random, what is the probability that the average length of the eight products in the sample is more than 22 inches?

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Solution:

If you wish to revise the concepts of Normal distribution, refer to Normal distribution curve and Normal distribution tables and areas

a)

Length of the product follows normal distribution ~ N (21.6, 1.12)

Objective: To find the probability of the randomly selected product falls outside the range of 20.5 inches to 21.5 inches = P(X<20.5) + P(X>21.5)

Convert x value to z (standard normal score) and then lookup for the area under the normal curve for calculated z values.

P(X<20.5) = P (Z< ((20.5 – mean)/standard deviation) = P (Z < (20.5 – 21.6)/1.1))

= P (Z < -1.00) = 0.1587

P(X>21.5) = P (Z> ((21.5 – mean)/standard deviation) = P (Z > (21.5 – 21.6)/1.1))

= P (Z > -0.091) = 0.5359 P(X<20.5) + P(X>21.5) = 0.1587 + 0.5359 = 0.6946

b)

Length of the product follows normal distribution ~ N (21.6, 1.12)

Objective: To find the probability that the sample mean will exceed 22 inches.

Concept:

Sample mean from a normally distributed population with mean µ and standard deviation σ will follow a normal distribution with mean µ and standard deviation σ / √n where n is the sample size.

Average length of the sample follows normal distribution ~ N (21.6, (1.1/√8)2)

P(X > 22) = P (Z > ((22 – mean)/standard deviation of the sample mean)

= P (Z > (22 – 21.6)/ (1.1/√8)) = P (Z > 0.13) = 0.4483 Normal distribution practice tests

Normal distribution problem 6

Question (Normal distribution problems): If it is given that the average stock returns on NYSE for the past one year is 4.3% and the standard deviation of returns is 1.9%, calculate the probability that the average returns over the past 1 year of 36 randomly selected stocks is not more than 5%?

Assume that the stock returns over the past one year are normally distributed.

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Solution:

Stock returns are normally distributed with mean of 4.3% and standard deviation of 1.9%.

We would like the calculate the probability of sample mean (Average of 36 stocks).

Concept:

Sample mean from a normally distributed population with mean µ and standard deviation σ will follow a normal distribution with mean µ and standard deviation σ / √n where n is the sample size.

Sample mean is normally distributed with mean 4.3% and standard deviation of 1.9%/6 = 0.32%

p(sample mean <=5%) = P(Z < = (5% – 4.3%)/0.32%) = p(Z <= 2.21)

P(Z<=2.21) = 0.9864 Normal distribution practice tests