A population of values has a normal distribution with $$ \mu=125.6 $$ and $$ \sigma=74 $$. You intend to draw a random sample of size $$ n=11 $$. Find the probability that a single randomly selected value is less than 83.2 . $$ P(x

Question

A population of values has a normal distribution with and . You intend to draw a random
sample of size .
Find the probability that a single randomly selected value is less than 83.2 .

Find the probability that a sample of size is randomly selected with a mean less than 83.2 .

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact -scores or -
scores rounded to 3 decimal places are accepted.
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Ask by Cole Hodges. in the United States

Dec 12,2024

UpStudy AI · Accepted Answer

To solve these problems, we'll use the properties of the normal distribution. Let's start with the first part:

### Part 1: Probability that a single randomly selected value is less than 83.2

We need to find $$ P(x < 83.2) $$. To do this, we'll convert the value 83.2 to a z-score, which represents the number of standard deviations the value is from the mean.

The formula for the z-score is:
$$ z = \frac{(x - \mu)}{\sigma} $$

Given:
- $$ \mu = 125.6 $$
- $$ \sigma = 74 $$
- $$ x = 83.2 $$

Let's calculate the z-score:
$$ z = \frac{(83.2 - 125.6)}{74} $$
$$ z = \frac{-42.4}{74} $$
$$ z \approx -0.573 $$

Now, we'll find the probability associated with this z-score using a standard normal distribution table or a calculator. The probability that a z-score is less than -0.573 is approximately 0.2810.

So, the probability that a single randomly selected value is less than 83.2 is:
$$ P(x < 83.2) \approx 0.2810 $$

### Part 2: Probability that a sample of size $$ n=11 $$ is randomly selected with a mean less than 83.2

We need to find $$ P(\bar{x} < 83.2) $$. The mean of a sample, $$\bar{x}$$, is also normally distributed with a mean equal to the population mean ($$\mu$$) and a standard deviation equal to the population standard deviation ($$\sigma$$) divided by the square root of the sample size ($$n$$).

The formula for the standard deviation of the sample mean is:
$$ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} $$

Given:
- $$ \mu = 125.6 $$
- $$ \sigma = 74 $$
- $$ n = 11 $$

Let's calculate the standard deviation of the sample mean:
$$ \sigma_{\bar{x}} = \frac{74}{\sqrt{11}} $$
$$ \sigma_{\bar{x}} \approx \frac{74}{3.3166} $$
$$ \sigma_{\bar{x}} \approx 22.410 $$

Now, we'll find the z-score for the sample mean of 83.2:
$$ z = \frac{(83.2 - 125.6)}{22.410} $$
$$ z = \frac{-42.4}{22.410} $$
$$ z \approx -1.895 $$

Using a standard normal distribution table or a calculator, the probability that a z-score is less than -1.895 is approximately 0.0287.

So, the probability that a sample of size $$ n=11 $$ is randomly selected with a mean less than 83.2 is:
$$ P(\bar{x} < 83.2) \approx 0.0287 $$

These are the solutions to the given problems, rounded to four decimal places as requested.
For the first part, the probability that a single randomly selected value is less than 83.2 is approximately 0.2810. For the second part, the probability that a sample of size $$ n=11 $$ is randomly selected with a mean less than 83.2 is approximately 0.0287.

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Solution

Part 1: Probability that a single randomly selected value is less than 83.2

Part 2: Probability that a sample of size is randomly selected with a mean less than 83.2

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