SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 98% confidence interval to 5 points, how many students should the administrator sample? Make sure to give a whole number answer.
Solution :
Given that,
standard deviation = = 300
margin of error = E = 5
At 98% confidence level the z is ,
= 1 - 98% = 1 - 0.98 = 0.02
/ 2 = 0.02 / 2 = 0.01
Z/2 = Z0.01 = 2.326
Sample size = n = ((Z/2 * ) / E)2
= ((2.326* 300) / 5)2
= 19476.9
Sample size = 19477
SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300....
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