
Normal Probability Distribution 
Question 1 Scores on a class exam have a mean of 85% and a standard
deviation of 5%.
Let x represents students test result on the exam (assume x is a random
normal variable).
(1) Draw a probability distribution of the student results
(2) What is the probability that a randomly selected student will get a grade of:
(a) Higher than 95% or less than 75%?
(b) Higher than the 85%?
(c) Between 90 and 95%?
(d) Lower than 90%?
(e) Between 75 and 85%?
(f) Within one standard deviation of the mean?
(g) Within two standard deviations of the man?
Solutions
(1) Probability Distribution of Exam Results:
mean = 85 and standard deviation = 5
Probability Calculation from program 
Question 2 Suppose z represents the Standard normal variable.
If a value is selected at random from the
z  distribution, find the probability that z is:
(a) less than 0
(b) Between 0.67 and 0.
(c) Between 2.3 and 1.45.
(d) Between 0.73 and 2.31.
(e) Less than 1.96
(g) Within one standard deviation of the mean
Solutions
From Standard Normal Table 
(a) Pr[<0] = Cum Pr[0] = 0.5
(b) Pr[0.67 to 0] = Pr[0]  Pr[0.67] = 0.5  0.2514 = 0.2486
(c) Pr[2.3 to 1.43] = Pr[1.43]  Pr[2.3] = 0.0764  0.0107 = 0.0655
(d) Pr[0.73 to 2.31] = Pr[2.31]  Pr[0.73] = 0.9896  0.2327 = 0.7569
(e) Pr[< 1.96] = Cum Pr[1.96] = 0.9750
(g) Pr[1 to 1] = Pr[1]  Pr[1] = 0.8413  0.1587 = 0.6826
Question 3 (6.21) A computer chip component is designed so that
they are replaced after 5 years.
A historical study of 10,000 parts over a 10 years period shows that
the mean time before first repair is
36 months and the standard deviation is 7 months.
(a) If the company plans to guarantee the chip component for 2 years,
what percent of the chip component will not satisfy the guarantee?
(b) The company plans to only repair 5% of all components, what should the guarantee period be?
Solutions
(a) So for 2 years or 24 months, zscore = (24  36) / 7 = 1.7143.., Guarantee not covered is
Cum Pr[z = 1.7143] = 0.0436 or 4.36%
(b) 5% repair equals a probability of 0.05 which correspond to zscore = 2.33
So z=2.33 = (x  48)/7, x = (7 x 2.33) + 48 = 31.69 months.
Question 4. (6.29) Assume that an experiment is a binomial experiment:
(a) Use the binomial table or probability distribution to find the probability
of x between 5 and 7, inclusive
where x is the number of defect, n = 15 trials and p = 0.6.
(b) Given the information in part (a) use the normal approximation to the binomial to find x between 5 and 7 inclusive.
Solutions
Binomial Pr[5 + 6 + 7 ] = 0.2038  Normal Approx. Pr[5 + 6 +
7] = 0.2038

Question 5 (6.41) A normal population has mean of 600 and standard
deviation of 300.
If the sample size is n= 100.
(a) Find the sample mean
(b) Find the sample standard deviation
(c) Find Pr[585 < x < 615]
(d) Find Pr[x > 600]
(e) If one value of x is selected, find Pr[585 < x < 615], compared with part (c)
(f) If one value of x is selected, find Pr[x > 600]
Solutions
(a) Sample mean is same of population mean = 600
(b) Sample standard deviation is (10)
(c) Pr[585 < x < 615], compute zscore (585600)/30 = 0.5 etc. So
Pr[0.5 < z < 0.5] = 0.6915  0.3085 = 0.3830
(d) Find Pr[x>630] , find zscore 530  500 / 30 = 1. Pr[ ] = 1  0.8413 = 0.1587
(e) If one value of x is selected, find Pr[585 < x < 615], compared with part (c)
Question 6. (6.57) let x be a normal random variable with mean 18 and standard deviation of 2. Find the following:
(a) The 10th percentile
(b) the 95th percentile
(c) The percent rank of x = 19
(d) The percent rank of x = 14.
Solutions
(a) The 10th percentile is z_{.010} = (x  mean) / std: z of 0.10 = 2.32 (where Pr = 0.10), mean = 18 and std = 2
So x = (2.32 x 2) + 18 = 13.36
(b) The 95 Th. percentile (z = 1.65 when Pr = 0.95), x = (1.65 x 2) + 18 = 21.3
(c) The percent rank of x = 19: z = (19  18) / 2 = 0.5, Pr[z=0.5] = 0.6915 or 69.15%
(d) The percent rank of x = 14 : z = (14  18) / 2 = 2, Pr[z=2] = 0.02275 or 2.28 %