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Forbes magazine reported at one time that the average number of weeks an individual is unemployed is 14.1 weeks. Assume that for the population of all unemployed individuals, the population mean length of unemployment is 14.1 weeks and that the population standard deviation is 4.4 weeks. Suppose you would like to select a random sample of 222 unemployed individuals for a follow-up study.

1. Find the probability that a single randomly selected value is between 14.3 and 14.5.

[tex]\[ P(14.3 \ \textless \ x \ \textless \ 14.5) = \square \][/tex]

2. Find the probability that a sample of size [tex]\( n = 222 \)[/tex] is randomly selected with a mean between 14.3 and 14.5.

[tex]\[ P(14.3 \ \textless \ \bar{x} \ \textless \ 14.5) = \square \][/tex]

Sagot :

GinnyAnswer
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]
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