Answer :
Sure! Let's solve the question step-by-step.
### Given Information:
- The population mean ([tex]\(\mu\)[/tex]) = 178.6
- The population standard deviation ([tex]\(\sigma\)[/tex]) = 99.3
- Sample size ([tex]\(n\)[/tex]) = 24
- Value to compare ([tex]\(x\)[/tex]) = 198.9
### Part (a): Find the probability a single randomly selected value is less than 198.9.
To find this probability, we will:
1. Calculate the z-score for the value 198.9.
2. Use the z-score to find the corresponding probability from the standard normal distribution.
#### Step 1: Calculate the z-score
The z-score is given by:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
Substitute the given values:
[tex]\[ z = \frac{198.9 - 178.6}{99.3} = \frac{20.3}{99.3} \approx 0.2044 \][/tex]
#### Step 2: Find the probability using the z-score
We can use standard normal distribution tables or a calculator to find the probability corresponding to [tex]\(z = 0.2044\)[/tex].
[tex]\[ P(x < 198.9) \approx 0.5810 \][/tex]
So,
[tex]\[ P(x < 198.9) \approx 0.5810 \][/tex]
### Part (b): Find the probability a sample of size [tex]\(n=24\)[/tex] has a mean less than 198.9.
For this part, we need to consider the sampling distribution of the sample mean. The mean of the sampling distribution ([tex]\(\mu_{\bar{x}}\)[/tex]) is the same as the population mean, and the standard deviation of the sampling distribution ([tex]\(\sigma_{\bar{x}}\)[/tex]) is given by:
[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]
#### Step 1: Calculate the standard deviation of the sample mean
[tex]\[ \sigma_{\bar{x}} = \frac{99.3}{\sqrt{24}} \approx 20.2612 \][/tex]
#### Step 2: Calculate the z-score for the sample mean
The z-score for the sample mean is given by:
[tex]\[ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \][/tex]
Substitute the given values:
[tex]\[ z = \frac{198.9 - 178.6}{20.2612} = \frac{20.3}{20.2612} \approx 1.0015 \][/tex]
#### Step 3: Find the probability using the z-score
Using standard normal distribution tables or a calculator to find the probability corresponding to [tex]\(z = 1.0015\)[/tex]:
[tex]\[ P(\bar{x} < 198.9) \approx 0.8417 \][/tex]
So,
[tex]\[ P(\bar{x} < 198.9) \approx 0.8417 \][/tex]
### Summary:
a. [tex]\( P(x < 198.9) \approx 0.5810 \)[/tex]
b. [tex]\( P(\bar{x} < 198.9) \approx 0.8417 \)[/tex]
### Given Information:
- The population mean ([tex]\(\mu\)[/tex]) = 178.6
- The population standard deviation ([tex]\(\sigma\)[/tex]) = 99.3
- Sample size ([tex]\(n\)[/tex]) = 24
- Value to compare ([tex]\(x\)[/tex]) = 198.9
### Part (a): Find the probability a single randomly selected value is less than 198.9.
To find this probability, we will:
1. Calculate the z-score for the value 198.9.
2. Use the z-score to find the corresponding probability from the standard normal distribution.
#### Step 1: Calculate the z-score
The z-score is given by:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
Substitute the given values:
[tex]\[ z = \frac{198.9 - 178.6}{99.3} = \frac{20.3}{99.3} \approx 0.2044 \][/tex]
#### Step 2: Find the probability using the z-score
We can use standard normal distribution tables or a calculator to find the probability corresponding to [tex]\(z = 0.2044\)[/tex].
[tex]\[ P(x < 198.9) \approx 0.5810 \][/tex]
So,
[tex]\[ P(x < 198.9) \approx 0.5810 \][/tex]
### Part (b): Find the probability a sample of size [tex]\(n=24\)[/tex] has a mean less than 198.9.
For this part, we need to consider the sampling distribution of the sample mean. The mean of the sampling distribution ([tex]\(\mu_{\bar{x}}\)[/tex]) is the same as the population mean, and the standard deviation of the sampling distribution ([tex]\(\sigma_{\bar{x}}\)[/tex]) is given by:
[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]
#### Step 1: Calculate the standard deviation of the sample mean
[tex]\[ \sigma_{\bar{x}} = \frac{99.3}{\sqrt{24}} \approx 20.2612 \][/tex]
#### Step 2: Calculate the z-score for the sample mean
The z-score for the sample mean is given by:
[tex]\[ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \][/tex]
Substitute the given values:
[tex]\[ z = \frac{198.9 - 178.6}{20.2612} = \frac{20.3}{20.2612} \approx 1.0015 \][/tex]
#### Step 3: Find the probability using the z-score
Using standard normal distribution tables or a calculator to find the probability corresponding to [tex]\(z = 1.0015\)[/tex]:
[tex]\[ P(\bar{x} < 198.9) \approx 0.8417 \][/tex]
So,
[tex]\[ P(\bar{x} < 198.9) \approx 0.8417 \][/tex]
### Summary:
a. [tex]\( P(x < 198.9) \approx 0.5810 \)[/tex]
b. [tex]\( P(\bar{x} < 198.9) \approx 0.8417 \)[/tex]
