Question

A manufacturer knows that their items have a normally distributed length, with a mean of 9.7 inches, and standard deviation of 3 inches.

(a) If one item is chosen at random, what is the probability that its length is less than 9.2 inches?

Answer:
Round to 4 decimal places.

(b) What are the mean and standard deviation of the sample mean length of 19 items?

Mean =


Standard deviation (standard error)
=
Leave the square root in your answer.

(c) If 19 items are chosen at random, what is the probability that their mean length is less than 9.2 inches?

Answer 1

To solve the problems, we will first make use of the Z-score formula and the properties of the normal distribution.

### (a) Probability that the length of one item is less than 9.2 inches

Given:
- Mean (\(\mu\)) = 9.7 inches
- Standard Deviation (\(\sigma\)) = 3 inches
- Value (X) = 9.2 inches

First, we calculate the Z-score:

\[
Z = \frac{X - \mu}{\sigma} = \frac{9.2 - 9.7}{3} = \frac{-0.5}{3} \approx -0.1667
\]

Next, we look up the Z-score in the standard normal distribution table or use a calculator to find the probability corresponding to \(Z = -0.1667\).

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

\[
P(Z < -0.1667) \approx 0.4336
\]

So, the probability that the length of one item is less than 9.2 inches is:

\[
\text{Probability} \approx 0.4336
\]

**Answer:** 0.4336

### (b) Mean and standard deviation of the sample mean length of 19 items:

1. The mean of the sample mean (\(\mu_{\bar{x}}\)) is the same as the population mean:

\[
\mu_{\bar{x}} = \mu = 9.7 \text{ inches}
\]

2. The standard deviation of the sample mean (also called standard error) is calculated using the formula:

\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{19}}
\]

**Mean = 9.7 inches**

**Standard deviation (standard error) = \(\frac{3}{\sqrt{19}}\)**

### (c) Probability that the mean length of 19 items is less than 9.2 inches

We need to calculate the Z-score for the sample mean:

1. Use the sample mean formula. Since \(n = 19\):

\[
Z = \frac{X - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{9.2 - 9.7}{\frac{3}{\sqrt{19}}}
\]

Calculating the Z-score:

\[
Z = \frac{-0.5}{\frac{3}{\sqrt{19}}} = \frac{-0.5 \cdot \sqrt{19}}{3}
\]

Let's use an approximate value for \(\sqrt{19}\):

\(\sqrt{19} \approx 4.3589\)

So,

\[
Z \approx \frac{-0.5 \cdot 4.3589}{3} \approx \frac{-2.17945}{3} \approx -0.7265
\]

Now, we find the probability corresponding to \(Z \approx -0.7265\):

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

\[
P(Z < -0.7265) \approx 0.2341
\]

So, the probability that the mean length of 19 items is less than 9.2 inches is:

**Answer:** 0.2341