You want to be able to estimate the average IQ of people in Winston-Salem. You would be satisfied knowing the average IQ plus-or-minus 5IQ points(using a 90% confidence interval). The standard deviation for IQ tests is typically 15 points; so you’ll use that as a guideline. How many people would you need to survey in order to be able to estimate the average to your desired degree of accuracy?

Sample size = n =25

Explanation:

standard deviation =s = \sigma =15

Margin of error = E = 5 At 90%

confidence level = 1 - 90% = 1 - 0.90 =0.10 /2 = 0.05

Z /2 = Z0.05 = 1.645

sample size n = [Z\alpha/2* \sigma / E] 2 n = ( 1.645*15 / 5)2 n =24.35

Sample size = n =25

fast by juice wrld

explanation:

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