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Sidana [21]
3 years ago
13

A population has a mean, μ = 89 and a standard deviation,σ = 24.

Mathematics
1 answer:
Cloud [144]3 years ago
3 0

Answer:

\bar X \sim N(\mu=89, \frac{24}{\sqrt{64}}=3)

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

Let X the random variable of interest for a population. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =89,\sigma =24)

We take a sample of n=64 . That represent the sample size.

The sample mean is defined as:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we find the expected value and variance for the sample mean we got:

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}[/tex]

The distribution for the sample mean is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=89, \frac{24}{\sqrt{64}}=3)

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