Select the correct number.

Raw scores on a certain test one year were normally distributed, with a mean of 156 and a standard deviation of 23. If 48,592 students took the test, about how many of the students scored less than 96?

\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline \multicolumn{8}{|c|}{Table shows values to the LEFT of the [tex][tex]$x$[/tex][/tex]-score} \\
\hline [tex][tex]$x$[/tex][/tex] & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 \\
\hline 2.5 & 0.99370 & 0.99396 & 0.99413 & 0.99430 & 0.99446 & 0.99461 & 0.99477 & 0.99492 \\
\hline 2.6 & 0.99534 & 0.99547 & 0.99560 & 0.99573 & 0.99585 & 0.99598 & 0.99609 & 0.99621 \\
\hline 2.7 & 0.99653 & 0.99664 & 0.99674 & 0.99683 & 0.99693 & 0.99702 & 0.99711 & 0.99720 \\
\hline 2.8 & 0.99744 & 0.99752 & 0.99760 & 0.99767 & 0.99774 & 0.99781 & 0.99788 & 0.99795 \\
\hline 2.9 & 0.99813 & 0.99819 & 0.99825 & 0.99831 & 0.99836 & 0.99841 & 0.99846 & 0.99851 \\
\hline -2.9 & 0.00187 & 0.00181 & 0.00175 & 0.00169 & 0.00164 & 0.00159 & 0.00154 & 0.00149 \\
\hline -2.8 & 0.00256 & 0.00248 & 0.00240 & 0.00233 & 0.00226 & 0.00219 & 0.00212 & 0.00205 \\
\hline -2.7 & 0.00347 & 0.00336 & 0.00326 & 0.00317 & 0.00307 & 0.00298 & 0.00289 & 0.00280 \\
\hline -2.6 & 0.00466 & 0.00453 & 0.00440 & 0.00427 & 0.00415 & 0.00402 & 0.00391 & 0.00379 \\
\hline -2.5 & 0.00621 & 0.00604 & 0.00587 & 0.00570 & 0.00554 & 0.00539 & 0.00523 & 0.00508 \\
\hline
\end{tabular}

A. 18,627
B. 60
C. 19,312
D. 220



Answer :

To find out approximately how many of the 48,592 students scored less than 96 on the test, we need to follow several steps using the properties of the normal distribution. Here is a detailed, step-by-step solution for this problem:

1. Identify the Given Parameters:
- Mean ([tex]\(\mu\)[/tex]): [tex]\(156\)[/tex]
- Standard deviation ([tex]\(\sigma\)[/tex]): [tex]\(23\)[/tex]
- Total number of students: [tex]\(48592\)[/tex]
- Raw score: [tex]\(96\)[/tex]

2. Calculate the Z-Score:
The Z-score helps us determine how many standard deviations a particular score is from the mean. For a score [tex]\(X\)[/tex] of [tex]\(96\)[/tex], the Z-score [tex]\(Z\)[/tex] is calculated using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]

Plugging in our values:
[tex]\[ Z = \frac{96 - 156}{23} = \frac{-60}{23} \approx -2.6087 \][/tex]

3. Find the Corresponding Probability:
The Z-score of [tex]\(-2.6087\)[/tex] indicates how far the score of 96 is from the mean in terms of standard deviations. We must now determine the probability of a student scoring below this Z-score using the standard normal distribution (Z-table).

Consulting the Z-table for values close to [tex]\(-2.6087\)[/tex], we find the corresponding cumulative probability. Based on Z-table values:
[tex]\[ P(Z = -2.6) \approx 0.00466 \][/tex]

4. Calculate the Number of Students Scoring Below 96:
Now, multiply the cumulative probability by the total number of students to find the number of students scoring below 96.
[tex]\[ \text{Number of students} = \text{Probability} \times \text{Total students} \][/tex]
[tex]\[ \text{Number of students} \approx 0.00466 \times 48592 \approx 226.44 \][/tex]

Since the number of students must be an integer, we round it to the nearest whole number:
[tex]\[ \text{Number of students} \approx 226 \][/tex]

5. Select the Correct Answer:
Based on our calculations, the closest option in the multiple-choice answers provided is:
[tex]\[ \boxed{220} \][/tex]

So, approximately 220 students scored less than 96 on the test.

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