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Answer :
Therefore, the probability that a sample of size `n = 202` is randomly selected with a mean between `35.4` and `42.8` is `0.0919`.
The mean is `μ = 35.9` and standard deviation is `σ = 65.4`.To find the probability that a single randomly selected value is between 35.4 and 42.8, the standardized value (z-score) for 35.4 and 42.8 is calculated as follows:
z1 = (35.4 - μ) / σ
= (35.4 - 35.9) / 65.4
= -0.0076z2 = (42.8 - μ) / σ
= (42.8 - 35.9) / 65.4
= 0.1058
Now, probability `P` (35.4 < x < 42.8) is given by:
P = P(z1 < z < z2)
Here, z-table for calculating `P(z1 < z < z2)`.
`P(z1 < z < z2) = 0.1299`.
Therefore, the probability that a single randomly selected value is between 35.4 and 42.8 is `0.1299`.
To find the probability that a sample of size `n = 202` is randomly selected with a mean between `35.4` and `42.8`. the mean of a sample follows a normal distribution with mean
Now, z-score for `x = 35.4` and `x = 42.8` are calculated as follows:
z1 = (35.4 - μ) / (σ / [tex]\sqrt{(n)}[/tex])
[tex]= (35.4 - 35.9) / (65.4 / \sqrt{(202)})[/tex]
= -1.3705z2
= (42.8 - μ) / (σ / [tex]\sqrt{(n)}[/tex])
[tex]= (42.8 - 35.9) / (65.4 / \sqrt{(202)})[/tex]
= 1.6584
Now, the probability `P` that the sample mean is between `35.4` and `42.8` is:
P = P(z1 < z < z2)
Here, use z-table for calculating `P(z1 < z < z2)`.
We get `P(z1 < z < z2) = 0.0919`.
To learn more about probability
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