Download The distribution of the annual incomes of a group of middle

The distribution of the annual incomes of a group of middle-management employees at Compton Plastics
approximates a normal distribution with a mean of $47,200 and a standard deviation of $800. (a) About 68 percent of
the incomes lie between what two amounts? (b) About 95 percent of the incomes lie between what two amounts? (c)
Virtually all of the incomes lie between what two amounts? (d) What are the median and the modal incomes? (e) Is
the distribution of incomes symmetrical?
μ = 47200, σ = 800
(a) 68% of the distribution lies between one standard deviation of the mean
 68% of the incomes lie between 47200 - 800 and 47200 + 800 = [$46,400, $48,000]
(b) 95% of the distribution lies between two standard deviations of the mean
 95% of the incomes lie between 47200 - 2(800) and 47200 + 2(800) = [$45,600, $48,800]
(c) Virtually all (99.73%) of the distribution lies between three standard deviations of the mean
Virtually all (99.73%) of the incomes lie between 47200- 3(800) and 47200 + 3(800) = [$44,800, $49,600]
(d) Since the distribution of incomes is normal, median = mode = mean = $47,200
(e) Yes, because normal distribution is symmetrical about the mean.
An analysis of the final test scores for Introduction to Business reveals the scores follow the normal probability
distribution. The mean of the distribution is 75 and the standard deviation is 8. The professor wants to award an A to
students whose score is in the highest 10 percent. What is the dividing point for those students who earn an A and
those earning a B?
μ = 75, σ = 8
z- score such that 10% of the distribution lies to its right is z = 1.2816
x = μ + z σ = 75 + 1.2816(8) = 85.25
The cut-off score is 85.25
A study by Great Southern Home Insurance revealed that none of the stolen goods were recovered by the
homeowners in 80 percent of reported thefts. (a) During a period in which 200 thefts occurred, what is the probability
that no stolen goods were recovered in 170 or more of the robberies? (b) During a period in which 200 thefts
occurred, what is the probability that no stolen goods were recovered in 150 or more robberies?
n = 200, p = 80% = 0.8, q = 1 - p = 0.2
μ = np = 200 * 0.8 = 160, σ = √(npq) = √(200 * 0.8 * 0.2) = 5.6569
(a) x = 169.5
z = (x - μ)/σ = (169.5 - 160)/5.6569 = 1.6794
P(x ≥ 170) = P(z > 1.6794) = 0.0465
Probability that no stolen goods were recovered in 170 or more of the robberies = 0.0465
(b) x = 149.5
z = (x - μ)/σ = (149.5 - 160)/5.6569 = -1.8561
P(x ≥ 150) = P(z > -1.8561) = 0.9683
Probability that no stolen goods were recovered in 150 or more of the robberies = 0.9683.