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

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

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.

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

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.

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).