A standard normal table is given to help answer the question.

\begin{tabular}{|c|c|}
\hline
[tex]$z$[/tex] & Probability \\
\hline
0.00 & 0.5000 \\
\hline
1.00 & 0.8413 \\
\hline
2.00 & 0.9772 \\
\hline
\end{tabular}



Answer :

Sure! Let's walk through the steps for solving the problem.

Problem: Suppose we have a sample size of 85, a population mean of 22, and a population standard deviation of 13. We want to find the probability that the sample mean falls between 19 and 23.

### Step 1: Determining the Z-scores
We first need to find the Z-scores corresponding to the bounds 19 and 23.

The formula for calculating a Z-score is:
[tex]\[ z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]

where:
- [tex]\( X \)[/tex] is the value for which we're finding the Z-score,
- [tex]\( \mu \)[/tex] is the population mean,
- [tex]\( \sigma \)[/tex] is the population standard deviation,
- [tex]\( n \)[/tex] is the sample size.

#### Lower Bound (19):
[tex]\[ z_{\text{lower}} = \frac{19 - 22}{\frac{13}{\sqrt{85}}} \][/tex]

#### Upper Bound (23):
[tex]\[ z_{\text{upper}} = \frac{23 - 22}{\frac{13}{\sqrt{85}}} \][/tex]

These calculations yield:
[tex]\[ z_{\text{lower}} = -2.1275871824522046 \][/tex]
[tex]\[ z_{\text{upper}} = 0.7091957274840682 \][/tex]

### Step 2: Finding Probabilities for Z-scores
Next, we use the Standard Normal Table (Z-table) to find the probabilities corresponding to these Z-scores.

The Standard Normal Table gives us the probability that a standard normal variable is less than or equal to a given value [tex]\( z \)[/tex].

Given Z-scores from the standard normal table:
- For [tex]\( z = 0.00 \)[/tex], the probability is 0.5000.
- For [tex]\( z = 1.00 \)[/tex], the probability is 0.8413.
- For [tex]\( z = 2.00 \)[/tex], the probability is 0.9772.

#### For [tex]\( z_{\text{upper}} = 0.71 \)[/tex]:

Using interpolation from the table, we can estimate the probability for [tex]\( z = 0.71 \)[/tex]:
[tex]\[ \approx 0.7611 \][/tex]

#### For [tex]\( z_{\text{lower}} = -2.13 \)[/tex]:

Since [tex]\( z = -2.13 \)[/tex] is between [tex]\(-2.00\)[/tex] and [tex]\(-2.50\)[/tex], by normal table, the probability for [tex]\( z = 2.13 \)[/tex] (considering symmetry):
[tex]\[ \approx 0.9834 \, (1 - 0.9834 = 0.0166) \rightarrow 0.0166 \][/tex]

### Step 3: Finding the Probability Between Bounds
To find the probability that the sample mean falls between the lower bound (19) and upper bound (23), we subtract the cumulative probability at [tex]\( z_{\text{lower}} \)[/tex] from the cumulative probability at [tex]\( z_{\text{upper}} \)[/tex]:

[tex]\[ \text{Probability} = P(Z < z_{\text{upper}}) - P(Z < z_{\text{lower}}) \][/tex]
[tex]\[ \text{Probability} = 0.7611 - 0.0166 \][/tex]
[tex]\[ \text{Probability} = 0.7445 \][/tex]

### Answer:
The probability that the sample mean falls between 19 and 23 is approximately 0.7445, indicating about a 74.45% chance of this event occurring.