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podryga [215]
3 years ago
11

A random variable x x has a Normal distribution with an unknown mean and a standard deviation of 12. Suppose that we take a rand

om sample of size n = 36 n=36 and find a sample mean of ¯ x = 98 x¯=98 . What is a 95% confidence interval for the mean of x x ?
Mathematics
1 answer:
Paul [167]3 years ago
8 0

Answer:

95% Confidence Interval: (94.08,101.92)    

Step-by-step explanation:

We are given the following in the question:

Sample mean, \bar{x} = 98

Sample size, n = 36

Alpha, α = 0.05

Population standard deviation, σ = 12

95% Confidence Interval:

Formula:

\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

98 \pm 1.96(\dfrac{12}{\sqrt{36}} ) = 98 \pm 3.92 = (94.08,101.92)

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3 years ago
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Step-by-step explanation:

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C.

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Hello,

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