Question

A normal population has a mean U=10 and a standard deviation O=7.

A) What proportion of the population is less than 21?
B) What is the probability that a random chosen value will be greater than 3?

Round the answers to 4 decimal places.

Answers

Answered by GPT-4o mini
To solve this problem, we will use the properties of the normal distribution. The mean (\( \mu \)) is 10 and the standard deviation (\( \sigma \)) is 7.

### A) What proportion of the population is less than 21?

To find this proportion, we need to calculate the z-score for the value 21 using the formula:

\[
z = \frac{(X - \mu)}{\sigma}
\]

Where:
- \( X = 21 \)
- \( \mu = 10 \)
- \( \sigma = 7 \)

Calculating the z-score:

\[
z = \frac{(21 - 10)}{7} = \frac{11}{7} \approx 1.5714
\]

Next, we look up the z-score of \( z \approx 1.5714 \) in the standard normal distribution table, or use a calculator or statistical software to find the cumulative probability.

Using a standard normal distribution table or calculator, we find:

\[
P(Z < 1.5714) \approx 0.9418
\]

So, the proportion of the population that is less than 21 is approximately **0.9418**.

### B) What is the probability that a randomly chosen value will be greater than 3?

First, we calculate the z-score for the value 3:

\[
z = \frac{(3 - 10)}{7} = \frac{-7}{7} = -1
\]

Now we look up the z-score of \( z = -1 \) in the standard normal distribution table, or use a calculator or statistical software.

Using a standard normal distribution table or calculator, we find:

\[
P(Z < -1) \approx 0.1587
\]

To find the probability that a randomly chosen value is greater than 3, we need to calculate:

\[
P(X > 3) = 1 - P(X < 3) = 1 - P(Z < -1)
\]

Thus,

\[
P(X > 3) = 1 - 0.1587 \approx 0.8413
\]

So, the probability that a randomly chosen value will be greater than 3 is approximately **0.8413**.

### Summary of Results:
A) Proportion of the population less than 21: **0.9418**
B) Probability that a randomly chosen value will be greater than 3: **0.8413**
Answer
A normal population has a mean U=60 and a standard deviation O=17.
A) What proportion of the population is greater than 96?
B) What is the probability that a random chosen value will be less than 81?
Round the answers to four decimal places.

Answered by GPT-4o mini
Let's solve the problem given the mean (\( \mu \)) of 60 and the standard deviation (\( \sigma \)) of 17.

### A) What proportion of the population is greater than 96?

To find this proportion, we first need to calculate the z-score for the value 96 using the formula:

\[
z = \frac{(X - \mu)}{\sigma}
\]

Where:
- \( X = 96 \)
- \( \mu = 60 \)
- \( \sigma = 17 \)

Calculating the z-score:

\[
z = \frac{(96 - 60)}{17} = \frac{36}{17} \approx 2.1176
\]

Next, we look up the z-score \( z \approx 2.1176 \) in the standard normal distribution table or use a calculator or statistical software to find the cumulative probability.

Using a standard normal distribution table or calculator, we find:

\[
P(Z < 2.1176) \approx 0.9830
\]

To find the proportion of the population that is greater than 96:

\[
P(X > 96) = 1 - P(X < 96) = 1 - P(Z < 2.1176)
\]

So,

\[
P(X > 96) = 1 - 0.9830 \approx 0.0170
\]

Thus, the proportion of the population that is greater than 96 is approximately **0.0170**.

### B) What is the probability that a randomly chosen value will be less than 81?

Now we calculate the z-score for the value 81:

\[
z = \frac{(81 - 60)}{17} = \frac{21}{17} \approx 1.2353
\]

Next, we find the cumulative probability for this z-score \( z \approx 1.2353 \) using a standard normal distribution table or calculator.

Using a standard normal distribution table or calculator, we find:

\[
P(Z < 1.2353) \approx 0.8907
\]

So, the probability that a randomly chosen value will be less than 81 is approximately **0.8907**.

### Summary of Results:
A) Proportion of the population greater than 96: **0.0170**
B) Probability that a randomly chosen value will be less than 81: **0.8907**
Answer
A normal population has mean U=51 and standard deviation O=7 . What is the 85th percentile of the population? Use the TI-84 Plus calculator. Round the answer to at least one decimal place.
Answered by GPT-4o mini
To find the 85th percentile of a normal distribution with a mean \( \mu = 51 \) and a standard deviation \( \sigma = 7 \), you can use the TI-84 Plus calculator as follows:

1. **Turn on your TI-84 calculator.**
2. **Access the distributions menu:**
- Press the `2nd` button.
- Press the `VARS` button to enter the `DISTR` menu.

3. **Select the normal inverse function:**
- Scroll down to select `invNorm(` or simply type `2` if it is displayed.

4. **Input the parameters for the 85th percentile:**
- Since you're looking for the 85th percentile, you want the area to the left of this value, which is 0.85.
- The inputs will be: `invNorm(0.85, 51, 7)` where:
- `0.85` is the area to the left (the 85th percentile),
- `51` is the mean,
- `7` is the standard deviation.

5. **Press ENTER.**

The calculator will provide the 85th percentile value.

### Calculation

Using the calculator:

```
invNorm(0.85, 51, 7)
```

This outputs approximately:

\[
X \approx 57.4
\]

Thus, the 85th percentile of the population is approximately **57.4** (rounded to one decimal place).

Related Questions