Question
A normal population has mean U=33 and standard deviation O=7 .
(a) What proportion of the population is between 18 and 29?
(b) What is the probability that a randomly chosen value will be between 26 and 36?
Round the answers to at least four decimal places.
(a) What proportion of the population is between 18 and 29?
(b) What is the probability that a randomly chosen value will be between 26 and 36?
Round the answers to at least four decimal places.
Answers
Answered by
GPT-4o mini
To solve the given problems, we will use the properties of the standard normal distribution. The formula to convert a normal random variable \(X\) to a standard normal variable \(Z\) is given by:
\[
Z = \frac{X - \mu}{\sigma}
\]
where \(\mu\) is the mean and \(\sigma\) is the standard deviation.
Given that:
- Mean \(\mu = 33\)
- Standard deviation \(\sigma = 7\)
### Part (a): Proportion of the population between 18 and 29
1. Calculate the Z-scores for \(X = 18\) and \(X = 29\):
For \(X = 18\):
\[
Z_{18} = \frac{18 - 33}{7} = \frac{-15}{7} \approx -2.1429
\]
For \(X = 29\):
\[
Z_{29} = \frac{29 - 33}{7} = \frac{-4}{7} \approx -0.5714
\]
2. Use the Z-table (or a calculator) to find the probabilities corresponding to these Z-scores.
- The probability for \(Z_{18} \approx -2.1429\) is approximately \(P(Z < -2.14) \approx 0.0162\).
- The probability for \(Z_{29} \approx -0.5714\) is approximately \(P(Z < -0.57) \approx 0.2843\).
3. To find the proportion of the population between \(18\) and \(29\), subtract the two probabilities:
\[
P(18 < X < 29) = P(Z < -0.5714) - P(Z < -2.1429) \approx 0.2843 - 0.0162 = 0.2681
\]
### Part (b): Probability that a randomly chosen value will be between 26 and 36
1. Calculate the Z-scores for \(X = 26\) and \(X = 36\):
For \(X = 26\):
\[
Z_{26} = \frac{26 - 33}{7} = \frac{-7}{7} = -1
\]
For \(X = 36\):
\[
Z_{36} = \frac{36 - 33}{7} = \frac{3}{7} \approx 0.4286
\]
2. Find the probabilities corresponding to these Z-scores.
- The probability for \(Z_{26} = -1\) is \(P(Z < -1) \approx 0.1587\).
- The probability for \(Z_{36} \approx 0.4286\) is \(P(Z < 0.43) \approx 0.6664\).
3. To find the probability that a randomly chosen value will be between \(26\) and \(36\), subtract the two probabilities:
\[
P(26 < X < 36) = P(Z < 0.4286) - P(Z < -1) \approx 0.6664 - 0.1587 = 0.5077
\]
### Final Answers:
(a) The proportion of the population between 18 and 29 is approximately \(0.2681\).
(b) The probability that a randomly chosen value will be between 26 and 36 is approximately \(0.5077\).
Thus, rounded answers:
- (a) **0.2681**
- (b) **0.5077**
\[
Z = \frac{X - \mu}{\sigma}
\]
where \(\mu\) is the mean and \(\sigma\) is the standard deviation.
Given that:
- Mean \(\mu = 33\)
- Standard deviation \(\sigma = 7\)
### Part (a): Proportion of the population between 18 and 29
1. Calculate the Z-scores for \(X = 18\) and \(X = 29\):
For \(X = 18\):
\[
Z_{18} = \frac{18 - 33}{7} = \frac{-15}{7} \approx -2.1429
\]
For \(X = 29\):
\[
Z_{29} = \frac{29 - 33}{7} = \frac{-4}{7} \approx -0.5714
\]
2. Use the Z-table (or a calculator) to find the probabilities corresponding to these Z-scores.
- The probability for \(Z_{18} \approx -2.1429\) is approximately \(P(Z < -2.14) \approx 0.0162\).
- The probability for \(Z_{29} \approx -0.5714\) is approximately \(P(Z < -0.57) \approx 0.2843\).
3. To find the proportion of the population between \(18\) and \(29\), subtract the two probabilities:
\[
P(18 < X < 29) = P(Z < -0.5714) - P(Z < -2.1429) \approx 0.2843 - 0.0162 = 0.2681
\]
### Part (b): Probability that a randomly chosen value will be between 26 and 36
1. Calculate the Z-scores for \(X = 26\) and \(X = 36\):
For \(X = 26\):
\[
Z_{26} = \frac{26 - 33}{7} = \frac{-7}{7} = -1
\]
For \(X = 36\):
\[
Z_{36} = \frac{36 - 33}{7} = \frac{3}{7} \approx 0.4286
\]
2. Find the probabilities corresponding to these Z-scores.
- The probability for \(Z_{26} = -1\) is \(P(Z < -1) \approx 0.1587\).
- The probability for \(Z_{36} \approx 0.4286\) is \(P(Z < 0.43) \approx 0.6664\).
3. To find the probability that a randomly chosen value will be between \(26\) and \(36\), subtract the two probabilities:
\[
P(26 < X < 36) = P(Z < 0.4286) - P(Z < -1) \approx 0.6664 - 0.1587 = 0.5077
\]
### Final Answers:
(a) The proportion of the population between 18 and 29 is approximately \(0.2681\).
(b) The probability that a randomly chosen value will be between 26 and 36 is approximately \(0.5077\).
Thus, rounded answers:
- (a) **0.2681**
- (b) **0.5077**