Question
A sample of size 115 will be drawn from a population with mean 48 and standard deviation 12. Use the TI-83 Plus/TI-84 Plus calculator.
(a) Find the probability that x will be between 45 and 50. Round the answer to at least four decimal places.
(a) Find the probability that x will be between 45 and 50. Round the answer to at least four decimal places.
Answers
GPT-4o mini
To find the probability that the sample mean \(\bar{x}\) falls between 45 and 50, we can use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normal if the sample size is sufficiently large.
Given:
- Population mean (\(\mu\)) = 48
- Population standard deviation (\(\sigma\)) = 12
- Sample size (n) = 115
First, we need to calculate the standard error (SE) of the sample mean:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{115}} \approx \frac{12}{10.7238} \approx 1.1173
\]
Now, we can find the z-scores for the sample mean values of 45 and 50:
1. For \(x = 45\):
\[
z_1 = \frac{x - \mu}{SE} = \frac{45 - 48}{1.1173} \approx \frac{-3}{1.1173} \approx -2.684
\]
2. For \(x = 50\):
\[
z_2 = \frac{x - \mu}{SE} = \frac{50 - 48}{1.1173} \approx \frac{2}{1.1173} \approx 1.7917
\]
Next, we will find the probabilities corresponding to these z-scores using the normal distribution table or a calculator.
Using the TI-83 Plus or TI-84 Plus calculator:
1. To find \(P(Z < -2.684)\):
- Use: `normalcdf(-1E99, -2.684)` to get the left tail probability.
2. To find \(P(Z < 1.7917)\):
- Use: `normalcdf(-1E99, 1.7917)` to get the left tail probability.
Finally, to find the probability that \(\bar{x}\) is between 45 and 50, we calculate:
\[
P(45 < \bar{x} < 50) = P(Z < 1.7917) - P(Z < -2.684)
\]
### Calculation Step-by-Step:
1. For \(P(Z < -2.684)\):
- Execute: `normalcdf(-1E99, -2.684)`
- Output: Approximately \(0.0036\)
2. For \(P(Z < 1.7917)\):
- Execute: `normalcdf(-1E99, 1.7917)`
- Output: Approximately \(0.9631\)
### Final Probability Calculation:
\[
P(45 < \bar{x} < 50) = 0.9631 - 0.0036 = 0.9595
\]
### Conclusion:
The probability that the sample mean \(\bar{x}\) will be between 45 and 50 is approximately **0.9595** when rounded to four decimal places.
Given:
- Population mean (\(\mu\)) = 48
- Population standard deviation (\(\sigma\)) = 12
- Sample size (n) = 115
First, we need to calculate the standard error (SE) of the sample mean:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{115}} \approx \frac{12}{10.7238} \approx 1.1173
\]
Now, we can find the z-scores for the sample mean values of 45 and 50:
1. For \(x = 45\):
\[
z_1 = \frac{x - \mu}{SE} = \frac{45 - 48}{1.1173} \approx \frac{-3}{1.1173} \approx -2.684
\]
2. For \(x = 50\):
\[
z_2 = \frac{x - \mu}{SE} = \frac{50 - 48}{1.1173} \approx \frac{2}{1.1173} \approx 1.7917
\]
Next, we will find the probabilities corresponding to these z-scores using the normal distribution table or a calculator.
Using the TI-83 Plus or TI-84 Plus calculator:
1. To find \(P(Z < -2.684)\):
- Use: `normalcdf(-1E99, -2.684)` to get the left tail probability.
2. To find \(P(Z < 1.7917)\):
- Use: `normalcdf(-1E99, 1.7917)` to get the left tail probability.
Finally, to find the probability that \(\bar{x}\) is between 45 and 50, we calculate:
\[
P(45 < \bar{x} < 50) = P(Z < 1.7917) - P(Z < -2.684)
\]
### Calculation Step-by-Step:
1. For \(P(Z < -2.684)\):
- Execute: `normalcdf(-1E99, -2.684)`
- Output: Approximately \(0.0036\)
2. For \(P(Z < 1.7917)\):
- Execute: `normalcdf(-1E99, 1.7917)`
- Output: Approximately \(0.9631\)
### Final Probability Calculation:
\[
P(45 < \bar{x} < 50) = 0.9631 - 0.0036 = 0.9595
\]
### Conclusion:
The probability that the sample mean \(\bar{x}\) will be between 45 and 50 is approximately **0.9595** when rounded to four decimal places.
(b) Find the 90th percentile of x . Round the answer to at least two decimal places.
GPT-4o mini
To find the 90th percentile of the sample mean \(\bar{x}\) from a population with mean \(\mu = 48\) and standard deviation \(\sigma = 12\), we can use the z-score corresponding to the 90th percentile and the standard error (SE) we calculated previously.
### Step 1: Find the z-score for the 90th percentile
Using a standard normal distribution table or a calculator, the z-score that corresponds to the 90th percentile is approximately **1.2816**.
### Step 2: Calculate the standard error (SE)
We already calculated the standard error in part (a):
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{115}} \approx 1.1173
\]
### Step 3: Use the z-score to find the 90th percentile of \(\bar{x}\)
The formula to find the sample mean at the 90th percentile is:
\[
x = \mu + z \cdot SE
\]
Substituting in the values we have:
\[
x = 48 + 1.2816 \cdot 1.1173
\]
### Step 4: Calculate \(x\)
\[
x = 48 + 1.2816 \cdot 1.1173 \approx 48 + 1.4324 \approx 49.4324
\]
### Step 5: Round to two decimal places
Rounding \(49.4324\) to two decimal places gives us:
\[
\text{90th percentile of } \bar{x} \approx 49.43
\]
### Final Answer:
The 90th percentile of the sample mean \(\bar{x}\) is approximately **49.43**.
### Step 1: Find the z-score for the 90th percentile
Using a standard normal distribution table or a calculator, the z-score that corresponds to the 90th percentile is approximately **1.2816**.
### Step 2: Calculate the standard error (SE)
We already calculated the standard error in part (a):
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{115}} \approx 1.1173
\]
### Step 3: Use the z-score to find the 90th percentile of \(\bar{x}\)
The formula to find the sample mean at the 90th percentile is:
\[
x = \mu + z \cdot SE
\]
Substituting in the values we have:
\[
x = 48 + 1.2816 \cdot 1.1173
\]
### Step 4: Calculate \(x\)
\[
x = 48 + 1.2816 \cdot 1.1173 \approx 48 + 1.4324 \approx 49.4324
\]
### Step 5: Round to two decimal places
Rounding \(49.4324\) to two decimal places gives us:
\[
\text{90th percentile of } \bar{x} \approx 49.43
\]
### Final Answer:
The 90th percentile of the sample mean \(\bar{x}\) is approximately **49.43**.