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A random variable ?x has a Normal distribution with an unknown mean and a standard deviation...

A random variable ?x has a Normal distribution with an unknown mean and a standard deviation of 12. Suppose that we take a random sample of size ?=36n=36 and find a sample mean of ?¯=98x¯=98 . What is a 95% confidence interval for the mean of ?x ?

(96.355,99.645)(96.355,99.645)

(97.347,98.653)(97.347,98.653)

(94.08,101.92)(94.08,101.92)

(74.48,121.52)

Math   Statistics-And-Probability   
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Answer #1

Solution :

Given that,

Z/2 = 1.96

Margin of error = E = Z/2* ( /n)

= 1.96 * (12 / 36)

= 3.92

At 95% confidence interval estimate of the population mean is,

- E < < + E

98 - 3.92 < < 98 + 3.92

94.08 < < 101.92

(94.08 , 101.92)

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