Question

Question 8: Daren and Josh are pretty good free throw shooters. Daren makes 75% of the free throws he attempts. Josh makes 80% of his free throws. Suppose we take separate random samples of 50 free throws each from Daren and Josh, and record the proportion of free throws that are made by each. Which of the following best describes the sampling distribution of pˆ_D − pˆ_J ?

A) Strong skew, with mean -0.05 and standard deviation 0.083

B) Approximately normal, with mean -0.05 and standard deviation 0.083

C) Shape cannot be determined, with mean of -0.05 and standard deviation 0.083

D) Strong skew, with mean -0.05 and standard deviation 0.118

E) Approximately normal, with mean -0.05 and standard deviation 0.118

Question 9: Donner Summit, California, is a popular ski resort area. Over the past 60 years, the annual snowfall totals of Donner Summit have followed a distribution that is strongly skewed right with a mean of 404 inches and a standard deviation of 129 inches. If many samples of size 9 were taken, which of the following would best describe the shape of the sampling distribution of X bar?

A) The shape is approximately normal since the sample size is reasonably large.

B) The shape is skewed right since np ≥ 10 and n(1 − p) ≥ 10 have not been met.

C) The shape is equally as skewed right as the population distribution.

D) The shape is skewed right but less so than the population distribution.

E) Cannot be determined from the given information.

Question 10: The heights of all adult males in Croatia are approximately normally distributed with a mean of 180 cm and a standard deviation of 7 cm. The heights of all adult females in Croatia are approximately normally distributed with a mean of 158 cm and a standard deviation of 9 cm. If independent random samples of 10 adult males and 10 adult females are taken, what is the probability that the difference in sample means (males – females) is greater than 20 cm?

A) 0.3463

B) 0.6537

C) 0.6827

D) 0.7104

E) 0.8687

Answer 1

1.

Mean of pˆD − pˆJ = 0.75 - 0.80 = -0.05

Standard deviation =

=

= 0.083

For large sample size,

B) Approximately normal, with mean -0.05 and standard deviation 0.083

2.

The sample is less than 30, so that

D) The shape is skewed right but less so than the population distribution.

3.

Mean difference = 180 - 158 = 22

Standard deviation of difference = =  3.605551

P( > 20) = P[Z > (20 - 22)/3.605551]

= P[Z > -0.5547]

= 0.7104

D) 0.7104