Question

A random sample of size 25 from a normal population has the mean 47.5 and the standard deviation 8.4. Does this information tend to support (or) refuse the claim that the mean of the population is 42.1?

Answer 1

Answer:

https://www.quora.com/In-a-normal-distribution-a-random-sample-of-size-25-specimens-has-a-mean-x-%CC%84-45-3-and-standard-deviation-s-7-9-Does-this-information-tend-to-support-or-refuse-the-claim-that-the-mean-of-the-population-is-40-5

Step-by-step explanation:

see this to get the answer

Answer 2

Given,

Size of sample = n = 25

Mean = 47.5

Standard deviation = s = 8.4

Claim that the mean of the population = μ = 42.1

To Find,

This information tends to support (or) refuse the claim of the mean=?

Solution,

From the formula of T-distribution,

$t =\frac{\bar{x}-\mu}{s / \sqrt{n}}$

putting values in this formula we get,

$t =\frac{{47.5}-42.1}{8.4 / \sqrt{25}}\\t =\frac{{47.5}-42.1}{8.4 / 5}\\\\t =\frac{{47.5}-42.1}{8.4 / 5}\\\\t =\frac{{5.4}}{8.4 / 5}$

t = 3.21

Hence, we conclude that this information tends to refuse the claim that the mean of the population is 42.1