Question

1. Consider a population proportion p = 0.27.

a. Calculate the standard error for the sampling distribution of the sample proportion when n = 17 and n = 65? (Round your final answer to 4 decimal places.)

b. Is the sampling distribution of the sample proportion approximately normal with n = 17 and n = 65?

c. Calculate the probability that the sample proportion is between 0.25 and 0.27 for n = 65. (Round "z-value" to 2 decimal places and final answer to 4 decimal places.)

2. The typical college student graduates with $27,300 in debt (The Boston Globe, May 27, 2012). Let debt among recent college graduates be normally distributed with a standard deviation of $6,000.

a. What is the probability that the average debt of two recent college graduates is more than $28,000? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

b. What is the probability that the average debt of two recent college graduates is more than $33,000? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

a) for n=17 ;

std error of proportion=σp=√(p*(1-p)/n)=

0.1077

for n=65

std error of proportion=σp=√(p*(1-p)/n)=

0.0551

b)

not in case of n=17 ;as np <5 ; therefore not approximately normal

for n=65 ; Yes as np>=5 and n(1-p) >=5

c)_

probability =

P(0.25<X<0.27)

=

P(-0.36<Z<0)=

0.5-0.3594=

0.1406

2)

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	27300
std deviation   =σ=	6000.0000
sample size       =n=	2
std error=σx̅=σ/√n=	4242.6407

probability =

P(X>28000)

=

P(Z>0.165)=

1-P(Z<0.16)=

1-0.5636=

0.4364

( please try 0.4345 if this comes wrong)

b)

probability =

P(X>33000)

=

P(Z>1.344)=

1-P(Z<1.34)=

1-0.9099=

0.0901

(try 0.0896 if this comes wrong)