Question

1. The Graduate Record Exam (GRE verbal test) has scores that are normally distributed with a mean of 532 and a standard deviation of 128. If one test taker is chosen at random, what is the probability that the person's GRE verbal score is above 500? Use Table 8.1 and leave answer in decimal form.

2. The Graduate Record Exam (GRE verbal test) has scores that are normally distributed with a mean of 532 and a standard deviation of 128. Suppose that a random sample of 22 people take the exam and their test scores are averaged. What is the probability that the mean test score for the sample is above 500? Leave answer in decimal form, rounded to 2 decimal places.

3. If the true proportion of men with red/green color blindness is 12%, and many random samples of 100 men were to be taken and a proportion of color blindness calculated for each sample, what would the standard deviation of the resulting sampling distribution of p-hat values be? Round your answer to 4 decimal places.

4. Assume the true proportion of men with red/green color blindness is 12%. A random sample of 100 men is tested to find a sample proportion of red/green color blindness. Use the Rule of Sample Proportions to find the interval in which such a sample proportion will fall 95% of the time.

Select one:

a. 8.75% to 15.25%

b. 5.75% to 18.25%

c. 5.5% to 18.5%

d. 6.25% to 17.75%

e. 7.75% to 16.25%

Answer 1

Solution:-

1) mean = = 532

standard deviation = = 128

P(x >500 ) = 1 - p( x< 500)

=1- p [(x - ) / < (500-532) /128 ]

=1- P(z < -0.25)

= 1 - 0.4013 = 0.5987

probability = 0.5987

2)

n = 22

=    = 532

= / n = 128/ 22 = 27.2897

P( > 500) = 1 - P( <500 )

= 1 - P[( - ) / < (500 -532) /27.2897 ]

= 1 - P(z < -1.17 )

= 1 - 0.121 = 0.8790

probabilty = 0.8790 = 0.88

3)

p = 0.12

1 - p = 1 -0.12 = 0.88

n = 100

= p =p = 0.12

=  [p( 1 - p ) / n] = [0.12*(0.88) /100 ] = 0.0325

4)

n = 100

Point estimate = sample proportion = = 0.120

1 - = 1-0.120 = 0.880

Z/2 = 1.96

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.960 * ((0.120*(0.880) /100 )

= 0.065

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.120 -0.065 < p < 0.120 +0.065

0.055 < p < 0.185

(0.055 ,0.185 ) = 5.5% to 18.5%

Answer = c. 5.5% to 18.5%