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Answer :
Sure! Let's work through the problem step by step.
We have a population with a normal distribution where the mean ([tex]\(\mu\)[/tex]) is 112.3 and the standard deviation ([tex]\(\sigma\)[/tex]) is 93.1. Let's find the probabilities as requested:
### Part a:
We want to find the probability that a single randomly selected value falls between 97.8 and 110.
1. Calculate the Z-scores for the lower and upper bounds. The Z-score is a way of standardizing a value by subtracting the mean and dividing by the standard deviation.
[tex]\[
Z_{\text{lower}} = \frac{97.8 - 112.3}{93.1}
\][/tex]
[tex]\[
Z_{\text{upper}} = \frac{110 - 112.3}{93.1}
\][/tex]
2. Use the standard normal distribution table (or a calculator with a normal distribution function) to find the probability associated with each Z-score.
3. Find the probability the random value is between these Z-scores:
[tex]\[
P(97.8 < X < 110) = P(Z_{\text{upper}}) - P(Z_{\text{lower}})
\][/tex]
After calculating, the probability is approximately 0.0520.
### Part b:
Now let's find the probability that the mean of a randomly selected sample of size [tex]\(n = 149\)[/tex] is between 97.8 and 110.
1. Calculate the standard error of the mean using the population standard deviation and sample size.
[tex]\[
\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{93.1}{\sqrt{149}}
\][/tex]
2. Calculate the Z-scores for the sample mean:
[tex]\[
Z_{\text{lower}} = \frac{97.8 - 112.3}{\text{Standard Error}}
\][/tex]
[tex]\[
Z_{\text{upper}} = \frac{110 - 112.3}{\text{Standard Error}}
\][/tex]
3. Use the standard normal distribution table (or a calculator) again to find the probabilities for these Z-scores.
4. Find the probability the sample mean is between these values:
[tex]\[
P(97.8 < M < 110) = P(Z_{\text{upper}}) - P(Z_{\text{lower}})
\][/tex]
After calculating, the probability is approximately 0.3529.
These calculations show the probability of different scenarios occurring given the normal distribution characteristics of the population and sample size.
We have a population with a normal distribution where the mean ([tex]\(\mu\)[/tex]) is 112.3 and the standard deviation ([tex]\(\sigma\)[/tex]) is 93.1. Let's find the probabilities as requested:
### Part a:
We want to find the probability that a single randomly selected value falls between 97.8 and 110.
1. Calculate the Z-scores for the lower and upper bounds. The Z-score is a way of standardizing a value by subtracting the mean and dividing by the standard deviation.
[tex]\[
Z_{\text{lower}} = \frac{97.8 - 112.3}{93.1}
\][/tex]
[tex]\[
Z_{\text{upper}} = \frac{110 - 112.3}{93.1}
\][/tex]
2. Use the standard normal distribution table (or a calculator with a normal distribution function) to find the probability associated with each Z-score.
3. Find the probability the random value is between these Z-scores:
[tex]\[
P(97.8 < X < 110) = P(Z_{\text{upper}}) - P(Z_{\text{lower}})
\][/tex]
After calculating, the probability is approximately 0.0520.
### Part b:
Now let's find the probability that the mean of a randomly selected sample of size [tex]\(n = 149\)[/tex] is between 97.8 and 110.
1. Calculate the standard error of the mean using the population standard deviation and sample size.
[tex]\[
\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{93.1}{\sqrt{149}}
\][/tex]
2. Calculate the Z-scores for the sample mean:
[tex]\[
Z_{\text{lower}} = \frac{97.8 - 112.3}{\text{Standard Error}}
\][/tex]
[tex]\[
Z_{\text{upper}} = \frac{110 - 112.3}{\text{Standard Error}}
\][/tex]
3. Use the standard normal distribution table (or a calculator) again to find the probabilities for these Z-scores.
4. Find the probability the sample mean is between these values:
[tex]\[
P(97.8 < M < 110) = P(Z_{\text{upper}}) - P(Z_{\text{lower}})
\][/tex]
After calculating, the probability is approximately 0.3529.
These calculations show the probability of different scenarios occurring given the normal distribution characteristics of the population and sample size.
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