Question

1. Suppose the proportion of the population in favor of a new law is 0.60.

(a) Suppose you take a sample of size n = 500. What is the probability that your sample proportion is 0.55 or lower?

(b) Suppose that you take a sample of size n = 400 and find the sample proportion ˆp = 0.70. How many standard deviations is ˆp from the mean of the distribution of all sample proportions?

2. A college senior who took the Graduate Record Examination (GRE) scored 647 on the verbal reasoning section and 654 on the quantitative reasoning section. The mean for the verbal reasoning section was 451 with a standard deviation of 120, and the mean score for the quantitative reasoning was 581 with standard deviation 153. Suppose that both distributions are nearly normal.

(a) What is the z-score for the verbal reasoning section?

(b) What is the z-score for the quantitative reasoning section?

(c) Which section did the student do better on relative to the other people that took the exam? Circle your answer.

A. The scores were the same

B. Verbal Reasoning

C. Quantitative Reasoning

(d) What is the percentile score for the verbal reasoning?

(e) What is the percentile score for the quantitative reasoning?

Answer 1

1a)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.6000
sample size       =n=	500
std error of proportion=σp=√(p*(1-p)/n)=	0.0219

probability =P(X<0.55)=(Z<0.55-0.6)/0.022)=P(Z<(-2.28)=0.0112

b)

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

0.0245

number of standard deviation =(0.7-0.60)/0.0245 = 4.0845

2)

a)

z score =(X-mean)/standard deviation=1.6333

b)

z score =(X-mean)/standard deviation=0.4771

c)

Verbal

d)

percentile score =P(Z<1.6333)=

94.8801

e)

percentile score =P(Z<0.4771) =

68.3363