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 = [tex]\frac{m-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
⇒ z =[tex]\frac{22-30}{\frac{8}{\sqrt{4} } }[/tex]
⇒ z = [tex]\frac{22-30}{\frac{8}{2} }[/tex]
⇒ z = [tex]-\frac{8}{4}[/tex]
⇒ 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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