1) Random samples of size 1,600 are drawn from a population in which the proport

Question

1) Random samples of size 1,600 are drawn from a population in which the proportion with the characteristic of interest is 0.05.   The sample size is large enough toassume that the sample proportion is normally distributed. True or False?

2) The proportion of a population with a characteristic of interest is p=0.82. The mean and standard deviation of the sample proportion pˆ obtained from random samples of size 900 are …. and …. (use 4 decimal points for the standard deviation)

3) According to data published by the U.S. Department of Agriculture, the average annual amount of chicken eaten by American adults (excluding vegetarians) is normally distributed with mean of 55 pounds and a standard deviation of 9.2 pounds.

(a) the probability that a randomly selected American adult eats at least one pound of chicken per week, that is, at least 52 pounds of chicken per year, is about

(a) 0.26

(b) 0.13

(c) 0.37

(d) 0.63

(e) 0.73

(b) Suppose an organization takes a sample of 50 American adults and records the annual chicken consumption of each. The probability that the sample mean would be 52 pounds or less is about

(a) 37%

(b) 10%

(c) 49%

(d) 51%

(e) 1%

4)The amount of time in hours spent each week watching television by children under twelve is normally distributed with mean 24.5 hours and standard deviation 6.23 hours. Five children under twelve are selected at random. The probability that the average number of hours that they spend each week watching television is at least 30 is about

Explanation / Answer

1)

True (As p=0.05 and n=1600 ; and np>=10 and n(1-p)>=10)

2)

mean of sample proportion=0.82

std deviation=sqrt(p(1-p)/n) =0.0128

3)

a) P(X>52)=P(Z>(52-55)/9.2)=P(Z>-0.33)=0.63

option D

b)

P(Xbar<52)=P(Z<(52-55)*sqrt(50)/9.2)=P(Z<-2.31)=1%

optio E

4)

P(Xbar>30)=P(Z>(30-24.5)*sqrt(5)/6.23)=P(Z>1.97)=0.0244

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