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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.
Thank you for reading the article A population of values has a normal distribution with tex mu 112 3 tex and tex sigma 93 1 tex a Find the probability that. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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