Answer :
Sure, let's walk through each part of the problem step by step.
### 1. Probability for a Single Randomly Selected Value
Firstly, we'll find the probability that a single random value from the population is between 14.3 and 14.5 weeks.
Given:
- Population mean, [tex]\( \mu = 14.1 \)[/tex]
- Population standard deviation, [tex]\( \sigma = 4.4 \)[/tex]
- Values of interest, [tex]\( x = 14.3 \)[/tex] and [tex]\( x = 14.5 \)[/tex]
We need to calculate the standard scores (z-scores) for these values. The z-score is given by the formula:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
For [tex]\( x = 14.3 \)[/tex]:
[tex]\[ z_{\text{lower}} = \frac{14.3 - 14.1}{4.4} = 0.04545454545454569 \][/tex]
For [tex]\( x = 14.5 \)[/tex]:
[tex]\[ z_{\text{upper}} = \frac{14.5 - 14.1}{4.4} = 0.09090909090909098 \][/tex]
Next, we find the probability associated with these z-scores using the standard normal distribution table (z-table).
The probability that a single randomly selected value is between 14.3 and 14.5 weeks (i.e., the area under the normal curve between these two z-scores) is:
[tex]\[ P(14.3 < x < 14.5) = P(z_{\text{lower}} < Z < z_{\text{upper}}) \approx 0.0181 \][/tex]
So the probability is approximately [tex]\( 0.0181 \)[/tex].
### 2. Probability for a Sample Mean
Now, we'll find the probability that the mean of a sample of 222 unemployed individuals falls between 14.3 and 14.5 weeks.
Given:
- Population mean, [tex]\( \mu = 14.1 \)[/tex]
- Population standard deviation, [tex]\( \sigma = 4.4 \)[/tex]
- Sample size, [tex]\( n = 222 \)[/tex]
- Values of interest, [tex]\( \bar{x} = 14.3 \)[/tex] and [tex]\( \bar{x} = 14.5 \)[/tex]
The standard error of the mean (SEM) is given by:
[tex]\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{4.4}{\sqrt{222}} \][/tex]
Next, we calculate the z-scores for the sample mean.
For [tex]\( \bar{x} = 14.3 \)[/tex]:
[tex]\[ z_{\text{lower\_sample}} = \frac{14.3 - 14.1}{\text{SEM}} \approx 0.6772574738977917 \][/tex]
For [tex]\( \bar{x} = 14.5 \)[/tex]:
[tex]\[ z_{\text{upper\_sample}} = \frac{14.5 - 14.1}{\text{SEM}} \approx 1.3545149477955774 \][/tex]
Using the standard normal distribution, we find the probability associated with these z-scores.
The probability that the sample mean of 222 unemployed individuals is between 14.3 and 14.5 weeks is:
[tex]\[ P(14.3 < \bar{x} < 14.5) = P(z_{\text{lower\_sample}} < Z < z_{\text{upper\_sample}}) \approx 0.1613 \][/tex]
So the probability is approximately [tex]\( 0.1613 \)[/tex].
### Summary of Results
Probability for a single randomly selected value:
[tex]\[ P(14.3 < x < 14.5) \approx 0.0181 \][/tex]
Probability for the sample mean (n=222):
[tex]\[ P(14.3 < \bar{x} < 14.5) \approx 0.1613 \][/tex]
### 1. Probability for a Single Randomly Selected Value
Firstly, we'll find the probability that a single random value from the population is between 14.3 and 14.5 weeks.
Given:
- Population mean, [tex]\( \mu = 14.1 \)[/tex]
- Population standard deviation, [tex]\( \sigma = 4.4 \)[/tex]
- Values of interest, [tex]\( x = 14.3 \)[/tex] and [tex]\( x = 14.5 \)[/tex]
We need to calculate the standard scores (z-scores) for these values. The z-score is given by the formula:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
For [tex]\( x = 14.3 \)[/tex]:
[tex]\[ z_{\text{lower}} = \frac{14.3 - 14.1}{4.4} = 0.04545454545454569 \][/tex]
For [tex]\( x = 14.5 \)[/tex]:
[tex]\[ z_{\text{upper}} = \frac{14.5 - 14.1}{4.4} = 0.09090909090909098 \][/tex]
Next, we find the probability associated with these z-scores using the standard normal distribution table (z-table).
The probability that a single randomly selected value is between 14.3 and 14.5 weeks (i.e., the area under the normal curve between these two z-scores) is:
[tex]\[ P(14.3 < x < 14.5) = P(z_{\text{lower}} < Z < z_{\text{upper}}) \approx 0.0181 \][/tex]
So the probability is approximately [tex]\( 0.0181 \)[/tex].
### 2. Probability for a Sample Mean
Now, we'll find the probability that the mean of a sample of 222 unemployed individuals falls between 14.3 and 14.5 weeks.
Given:
- Population mean, [tex]\( \mu = 14.1 \)[/tex]
- Population standard deviation, [tex]\( \sigma = 4.4 \)[/tex]
- Sample size, [tex]\( n = 222 \)[/tex]
- Values of interest, [tex]\( \bar{x} = 14.3 \)[/tex] and [tex]\( \bar{x} = 14.5 \)[/tex]
The standard error of the mean (SEM) is given by:
[tex]\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{4.4}{\sqrt{222}} \][/tex]
Next, we calculate the z-scores for the sample mean.
For [tex]\( \bar{x} = 14.3 \)[/tex]:
[tex]\[ z_{\text{lower\_sample}} = \frac{14.3 - 14.1}{\text{SEM}} \approx 0.6772574738977917 \][/tex]
For [tex]\( \bar{x} = 14.5 \)[/tex]:
[tex]\[ z_{\text{upper\_sample}} = \frac{14.5 - 14.1}{\text{SEM}} \approx 1.3545149477955774 \][/tex]
Using the standard normal distribution, we find the probability associated with these z-scores.
The probability that the sample mean of 222 unemployed individuals is between 14.3 and 14.5 weeks is:
[tex]\[ P(14.3 < \bar{x} < 14.5) = P(z_{\text{lower\_sample}} < Z < z_{\text{upper\_sample}}) \approx 0.1613 \][/tex]
So the probability is approximately [tex]\( 0.1613 \)[/tex].
### Summary of Results
Probability for a single randomly selected value:
[tex]\[ P(14.3 < x < 14.5) \approx 0.0181 \][/tex]
Probability for the sample mean (n=222):
[tex]\[ P(14.3 < \bar{x} < 14.5) \approx 0.1613 \][/tex]
