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Nataly_w [17]
3 years ago
5

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 n = 36 and find a sample mean of ¯ x = 98 . What is a 95% confidence interval for the mean of x ?
Mathematics
1 answer:
Zielflug [23.3K]3 years ago
6 0

Answer: The 95% confidence interval for the mean of x is (94.08, 101.92) .

Step-by-step explanation:

We are given that ,

A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.

i.e. \sigma= 12

Also, it is given that , Sample mean \overline{x}=98 having sample size : n= 36

For 95% confidence ,

Significance level : \alpha=1-0.95=0.05

By using the z-value table , the two-tailed critical value for 95% Confidence interval :

z_{\alpha/2}=z_{0.025}=1.96

We know that the confidence interval for unknown population mean(\mu) is given by :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}

, where \overline{x} = Sample mean

\sigma = Population standard deviation

z_{\alpha/2} = Critical z-value.

Substitute all the given values, then the required confidence interval will be :

98\pm (1.96)\dfrac{12}{\sqrt{36}}

=98\pm (1.96)\dfrac{12}{6}

=98\pm (1.96)(2)

=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)

Therefore, the 95% confidence interval for the mean of x is (94.08, 101.92) .

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