Answer :
To solve this problem, we need to find the probability that a sample of size [tex]\( n = 176 \)[/tex] is randomly selected with a mean less than 248, given that the population has a normal distribution with mean [tex]\(\mu = 234.7\)[/tex] and standard deviation [tex]\(\sigma = 98.2\)[/tex].
Here is the step-by-step approach:
1. Calculate the Standard Error of the Mean (SEM):
The standard error of the mean is calculated using the formula:
[tex]\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\][/tex]
where [tex]\(\sigma = 98.2\)[/tex] is the population standard deviation and [tex]\(n = 176\)[/tex] is the sample size.
[tex]\[
\text{SEM} = \frac{98.2}{\sqrt{176}} \approx 7.4021
\][/tex]
2. Calculate the Z-score for the Sample Mean:
The Z-score tells us how many standard deviations away the sample mean is from the population mean. It is calculated using the formula:
[tex]\[
Z = \frac{\bar{X} - \mu}{\text{SEM}}
\][/tex]
where [tex]\(\bar{X} = 248\)[/tex] is the sample mean we are interested in, [tex]\(\mu = 234.7\)[/tex] is the population mean, and [tex]\(\text{SEM} \approx 7.4021\)[/tex] is the standard error.
[tex]\[
Z = \frac{248 - 234.7}{7.4021} \approx 1.7968
\][/tex]
3. Find the Probability:
Now, we look for the probability that the sample mean is less than 248. This is equivalent to finding the cumulative probability up to the Z-score we just calculated. We use the standard normal distribution table or a calculator to find this probability:
[tex]\[
P(M < 248) \approx 0.9638
\][/tex]
Thus, the probability that a sample of size 176 has a mean less than 248 is approximately [tex]\(0.9638\)[/tex].
Here is the step-by-step approach:
1. Calculate the Standard Error of the Mean (SEM):
The standard error of the mean is calculated using the formula:
[tex]\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\][/tex]
where [tex]\(\sigma = 98.2\)[/tex] is the population standard deviation and [tex]\(n = 176\)[/tex] is the sample size.
[tex]\[
\text{SEM} = \frac{98.2}{\sqrt{176}} \approx 7.4021
\][/tex]
2. Calculate the Z-score for the Sample Mean:
The Z-score tells us how many standard deviations away the sample mean is from the population mean. It is calculated using the formula:
[tex]\[
Z = \frac{\bar{X} - \mu}{\text{SEM}}
\][/tex]
where [tex]\(\bar{X} = 248\)[/tex] is the sample mean we are interested in, [tex]\(\mu = 234.7\)[/tex] is the population mean, and [tex]\(\text{SEM} \approx 7.4021\)[/tex] is the standard error.
[tex]\[
Z = \frac{248 - 234.7}{7.4021} \approx 1.7968
\][/tex]
3. Find the Probability:
Now, we look for the probability that the sample mean is less than 248. This is equivalent to finding the cumulative probability up to the Z-score we just calculated. We use the standard normal distribution table or a calculator to find this probability:
[tex]\[
P(M < 248) \approx 0.9638
\][/tex]
Thus, the probability that a sample of size 176 has a mean less than 248 is approximately [tex]\(0.9638\)[/tex].
