Intelligence scores. For people in any age group, intelligence scores are distri

Question

Intelligence scores. For people in any age group, intelligence scores are distributed approximately normally with µ = 100 and = 15. In each part of this question, use two places after the decimal point for z scores and four places after the decimal point for proportions. For each part (except part e), include an appropriate sketch. IN EACH PART, MAKE CLEAR WHAT YOUR ANSWER IS!!! (For all parts that involve distributions of sample means—parts b, c, d, f, g, h, and i—you have to calculate and use the standard deviation of a distribution of means for samples of the specified size.)

a) What proportion of intelligence scores is higher than 105?

b) What proportion of means of random samples of 9 intelligence scores is higher than 105?

c) What proportion of means of random samples of 25 intelligence scores is higher than 105?

What proportion of means of random samples of 100 intelligence scores is higher than 105?

d) Correct answers to parts a-d illustrate the Law of Large Numbers. Describe the pattern of results for parts a-d by filling each blank in the following statement with the word increases or decreases. (Note: If your answers to parts a-d don’t follow the pattern you describe, you should probably check your answers.) As the sample size _______________________, the proportion of means of random samples at least some distance from µ ____________________________ .

Explanation / Answer

a) P(proportion of intelligence scores is higher than 105) =P(X>105)=P(Z>(105-100)/15)=P(Z>0.3333)=0.3694

b)

std error =std deviaiton/(n)1/2 =15/(9)1/2 =5

P(proportion of intelligence scores is higher than 105) =P(Xbar>105)=P(Z>(105-100)/5)=P(Z>1)=0.1587

c)

std error =std deviaiton/(n)1/2 =15/(25)1/2 =3

P(proportion of intelligence scores is higher than 105) =P(Xbar>105)=P(Z>(105-100)/3)=P(Z>1.667)=0.0.0478

for n=100

std error =std deviaiton/(n)1/2 =15/(100)1/2 =1.5

P(proportion of intelligence scores is higher than 105) =P(Xbar>105)=P(Z>(105-100)/1.5)=P(Z>3.333)=0.0004

d) As the sample size increases the proportion of means of random samples at least some distance from µ decreases/

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