A population of values has a normal distribution with [tex]\(\mu = 152.3\)[/tex] and [tex]\(\sigma = 63.6\)[/tex]. If a random sample of size [tex]\(n = 23\)[/tex] is selected:

a. Find the probability that a single randomly selected value is less than 144.3. Round your answer to four decimals.

[tex]\[ P(X \ \textless \ 144.3) = \ \square \][/tex]

b. Find the probability that a sample of size [tex]\(n = 23\)[/tex] is randomly selected with a mean less than 144.3. Round your answer to four decimals.

[tex]\[ P(M \ \textless \ 144.3) = \ \square \][/tex]



Answer :

Let's walk through each of the parts step by step:

### Part (a)
We need to find the probability that a single randomly selected value from this population is less than 144.3.

Given the parameters of the normal distribution:
- Mean ([tex]\(\mu\)[/tex]) = 152.3
- Standard Deviation ([tex]\(\sigma\)[/tex]) = 63.6

First, we calculate the z-score for the value 144.3 using the formula for the z-score:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
where [tex]\(x = 144.3\)[/tex].

Plugging in the values:
[tex]\[ z = \frac{144.3 - 152.3}{63.6} = -0.125786 \][/tex]

Next, we use the z-score to find the probability [tex]\(P(X < 144.3)\)[/tex]. This probability is the cumulative distribution function (CDF) of the standard normal distribution evaluated at [tex]\(z = -0.1258\)[/tex].

The probability that a single randomly selected value is less than 144.3 is:
[tex]\[ P(X < 144.3) = 0.4500 \][/tex]

### Part (b)
We need to find the probability that the sample mean of a sample of size [tex]\( n = 23 \)[/tex] is less than 144.3.

For the sampling distribution of the sample mean, the mean ([tex]\(\mu_M\)[/tex]) is the same as the population mean, 152.3, and the standard deviation ([tex]\(\sigma_M\)[/tex]) is given by:
[tex]\[ \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{63.6}{\sqrt{23}} = 13.2615167 \][/tex]

Next, we calculate the z-score for the sample mean:
[tex]\[ z = \frac{M - \mu_M}{\sigma_M} \][/tex]
where [tex]\( M = 144.3 \)[/tex].

Plugging in the values:
[tex]\[ z = \frac{144.3 - 152.3}{13.3} = -0.603249 \][/tex]

Finally, we find the probability [tex]\(P(M < 144.3)\)[/tex] using the z-score:
[tex]\[ P(M < 144.3) = 0.2732 \][/tex]

### Final Results:
- a. The probability that a single randomly selected value is less than 144.3 is [tex]\( P(X < 144.3) = 0.4500 \)[/tex].
- b. The probability that the sample mean of a sample of size 23 is less than 144.3 is [tex]\( P(M < 144.3) = 0.2732 \)[/tex].