Question

- A student wants to estimate the average mark on a statistics test to within 5 marks (out of 100), with 95% confidence. A preliminary sample yielded a standard deviation of 15.54. How big a sample size is necessary? Round to the nearest whole number.

Answer #1

sampling error , E = 5

Confidence Level , CL= 95%

alpha = 1-CL = 5%

Z value = Zα/2 = 1.960 [excel formula
=normsinv(α/2)]

Sample Size,n = (Z*σ/E)² = (1.96*15.54/5)² =
37.107

S**o,Sample Size needed=
38**

the average mark (in percentage) in statistics test for a class
comprising 300 students is 54%. if the marks are normally
distributed with a standard deviation of 8%, the number of students
with a mark of over 70% is

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