Question

A random sample of n = 4 scores is obtained from a normal population with m = 30 and s = 8. what is the probability that the sample mean will be smaller than m = 22?

Answer 1

the probability that the sample mean will be smaller than m = 22 is 0.02275.

For the given question,

A z-score measures exactly how many standard deviations a data point is above or below the mean. It allows us to calculate the probability of a score occurring within our normal distribution and enables us to compare two scores that are from different normal distributions.

Here sample size, n = 4

mean, μ = 30

standard deviation, σ = 8

The probability that the sample mean will be smaller than m = 22 is

z = $\frac{m-\mu}{\frac{\sigma}{\sqrt{n} } }$

⇒ z =$\frac{22-30}{\frac{8}{\sqrt{4} } }$

⇒ z = $\frac{22-30}{\frac{8}{2} }$

⇒ z = $-\frac{8}{4}$

⇒ z = -2

Refer the Z table for p value,

Thus for P(z < -2) is 0.02275.

Hence we can conclude that the probability that the sample mean will be smaller than m = 22 is 0.02275.

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