Question
A sample of size 21 will be drawn from a population with mean 5 and standard deviation 3.
(a) Is it appropriate to use the normal distribution to find probabilities for x?
(b) If appropriate find the probability that x will be greater than 4.
(c) If appropriate find the 30th percentile of x.
(a) Is it appropriate to use the normal distribution to find probabilities for x?
(b) If appropriate find the probability that x will be greater than 4.
(c) If appropriate find the 30th percentile of x.
Answers
GPT-4o mini
To solve the questions, we first need to understand the underlying principles of sampling distributions and the Central Limit Theorem.
### (a) Is it appropriate to use the normal distribution to find probabilities for \( \bar{x} \)?
In this case, we are drawing a sample of size \( n = 21 \) from a population with a known mean \( \mu = 5 \) and standard deviation \( \sigma = 3 \).
The Central Limit Theorem states that the sampling distribution of the sample mean \( \bar{x} \) will be approximately normally distributed if the sample size is large enough (generally \( n \geq 30 \) is considered sufficient). However, for smaller samples, if the underlying population is normally distributed, the sample mean distribution will also be normally distributed.
Since we do not have any information about the shape of the underlying population distribution, we cannot definitively say whether it is appropriate to use the normal distribution. If we assume the population is normal (or approximately normal), then yes, we can use the normal distribution.
However, if the population is not normal, the sample size of 21 may be too small for the normal approximation to be valid.
**Conclusion for (a)**: It is only appropriate to use the normal distribution if we assume that the population is normally distributed. Otherwise, it may not be appropriate.
---
### (b) If appropriate, find the probability that \( \bar{x} \) will be greater than 4.
If we proceed with the assumption that we can use the normal distribution, we can find the sample mean's standard deviation (also called the standard error) using the formula:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{21}} \approx 0.653
\]
Next, we standardize the value (convert to a Z-score):
\[
Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{4 - 5}{0.653} \approx -1.527
\]
Now we can look up the Z-score in the standard normal distribution tables or use a calculator to find the probability.
Using standard normal distribution tables or a calculator:
\[
P(Z > -1.527) \approx 1 - P(Z < -1.527) \approx 1 - 0.0634 = 0.9366
\]
**Conclusion for (b)**: The probability that \( \bar{x} \) will be greater than 4 is approximately \( 0.9366 \).
---
### (c) If appropriate, find the 30th percentile of \( \bar{x} \).
To find the 30th percentile, we need the Z-score for the 30th percentile, which can be found in the standard normal distribution table. The Z-score for the 30th percentile is approximately \( -0.524 \).
Now, we can convert this Z-score back to the value of \( \bar{x} \):
\[
\bar{x} = \mu + Z \cdot \sigma_{\bar{x}} = 5 + (-0.524) \cdot 0.653 \approx 5 - 0.342 = 4.658
\]
**Conclusion for (c)**: The 30th percentile of \( \bar{x} \) is approximately \( 4.658 \).
### Summary:
- (a) Using the normal distribution depends on the underlying population's distribution.
- (b) The probability that \( \bar{x} > 4 \) is approximately \( 0.9366 \) (assuming normality).
- (c) The 30th percentile of \( \bar{x} \) is approximately \( 4.658 \) (assuming normality).
### (a) Is it appropriate to use the normal distribution to find probabilities for \( \bar{x} \)?
In this case, we are drawing a sample of size \( n = 21 \) from a population with a known mean \( \mu = 5 \) and standard deviation \( \sigma = 3 \).
The Central Limit Theorem states that the sampling distribution of the sample mean \( \bar{x} \) will be approximately normally distributed if the sample size is large enough (generally \( n \geq 30 \) is considered sufficient). However, for smaller samples, if the underlying population is normally distributed, the sample mean distribution will also be normally distributed.
Since we do not have any information about the shape of the underlying population distribution, we cannot definitively say whether it is appropriate to use the normal distribution. If we assume the population is normal (or approximately normal), then yes, we can use the normal distribution.
However, if the population is not normal, the sample size of 21 may be too small for the normal approximation to be valid.
**Conclusion for (a)**: It is only appropriate to use the normal distribution if we assume that the population is normally distributed. Otherwise, it may not be appropriate.
---
### (b) If appropriate, find the probability that \( \bar{x} \) will be greater than 4.
If we proceed with the assumption that we can use the normal distribution, we can find the sample mean's standard deviation (also called the standard error) using the formula:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{21}} \approx 0.653
\]
Next, we standardize the value (convert to a Z-score):
\[
Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{4 - 5}{0.653} \approx -1.527
\]
Now we can look up the Z-score in the standard normal distribution tables or use a calculator to find the probability.
Using standard normal distribution tables or a calculator:
\[
P(Z > -1.527) \approx 1 - P(Z < -1.527) \approx 1 - 0.0634 = 0.9366
\]
**Conclusion for (b)**: The probability that \( \bar{x} \) will be greater than 4 is approximately \( 0.9366 \).
---
### (c) If appropriate, find the 30th percentile of \( \bar{x} \).
To find the 30th percentile, we need the Z-score for the 30th percentile, which can be found in the standard normal distribution table. The Z-score for the 30th percentile is approximately \( -0.524 \).
Now, we can convert this Z-score back to the value of \( \bar{x} \):
\[
\bar{x} = \mu + Z \cdot \sigma_{\bar{x}} = 5 + (-0.524) \cdot 0.653 \approx 5 - 0.342 = 4.658
\]
**Conclusion for (c)**: The 30th percentile of \( \bar{x} \) is approximately \( 4.658 \).
### Summary:
- (a) Using the normal distribution depends on the underlying population's distribution.
- (b) The probability that \( \bar{x} > 4 \) is approximately \( 0.9366 \) (assuming normality).
- (c) The 30th percentile of \( \bar{x} \) is approximately \( 4.658 \) (assuming normality).