To determine the number of students scoring less than 96 on the standardized test, follow these steps:
1. Calculate the z-score for the score of 96 using the formula:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
where [tex]\( X \)[/tex] is the score, [tex]\( \mu \)[/tex] is the mean, and [tex]\( \sigma \)[/tex] is the standard deviation.
Given:
[tex]\[
X = 96, \quad \mu = 156, \quad \sigma = 23
\][/tex]
Plugging in the values:
[tex]\[
z = \frac{96 - 156}{23} = \frac{-60}{23} \approx -2.6087
\][/tex]
2. Find the probability corresponding to the z-score of [tex]\(-2.6087\)[/tex] from the standard normal distribution table.
- Locate [tex]\(-2.60\)[/tex] in the row for [tex]\(-2.6\)[/tex].
- Look across the columns to find the closest value for the hundredths place, which is approximately [tex]\(-2.61\)[/tex].
Based on the standard normal distribution table provided, the probability for a z-score [tex]\(-2.6\)[/tex] is approximately 0.00466.
3. Calculate the number of students scoring less than 96 using the probability:
[tex]\[
\text{Number of students} = \text{Probability} \times \text{Total number of students}
\][/tex]
Given:
[tex]\[
\text{Probability} = 0.00466, \quad \text{Total number of students} = 48592
\][/tex]
Plugging in the values:
[tex]\[
\text{Number of students} = 0.00466 \times 48592 \approx 226.43872
\][/tex]
Therefore, approximately 226 students scored less than 96 on the standardized test.
