It is known that the population variance equals 529.With a .95 probability, the sample size that needs to be taken if the desired margin of error is 4 or less is

A) 508.
B) 127.
C) 509.
D) 128.

Question

It is known that the population variance equals 529.With a .95 probability, the sample size that needs to be taken if the desired margin of error is 4 or less isA)  508.B)  127.C)  509.D)  128.

Sathvik Ponnam · Accepted Answer

Final Answer : D, Explanation :  The formula for determining the sample size needed to find a desired margin of error is as follows:n = (z^2 * σ^2) / E^2Where:n = sample sizez = z-score for desired level of confidence (in this case, .95 corresponds to a z-score of 1.96)σ^2 = population variance (given as 529)E = desired margin of error (given as 4)Plugging in the values, we get:n = (1.96^2 * 529) / 4^2n = 128.15Rounding up to the nearest integer, we get a required sample size of 128. Therefore, the answer is D.

It is known that the population variance equals 529.With a .95 probability, the sample size that needs to be taken if the desired margin of error is 4 or less is

A) 508.
B) 127.
C) 509.
D) 128.

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