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Raw scores on a certain standardized 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?

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline \multicolumn{8}{|c|}{Table shows values to the LEFT of the z-score} \\
\hline \multicolumn{1}{|c|}{} & [tex]0.00[/tex] & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 \\
\hline [tex]2.5[/tex] & 0.99379 & 0.99396 & 0.99413 & 0.99430 & 0.99446 & 0.99461 & 0.99477 & 0.99492 \\
\hline [tex]2.6[/tex] & 0.99534 & 0.99547 & 0.99560 & 0.99573 & 0.99585 & 0.99598 & 0.99609 & 0.99621 \\
\hline [tex]2.7[/tex] & 0.99653 & 0.99664 & 0.99674 & 0.99683 & 0.99693 & 0.99702 & 0.99711 & 0.99720 \\
\hline [tex]2.8[/tex] & 0.99744 & 0.99752 & 0.99760 & 0.99767 & 0.99774 & 0.99781 & 0.99788 & 0.99795 \\
\hline [tex]2.9[/tex] & 0.99813 & 0.99819 & 0.99825 & 0.99831 & 0.99836 & 0.99841 & 0.99846 & 0.99851 \\
\hline [tex]-2.9[/tex] & 0.00187 & 0.00181 & 0.00175 & 0.00169 & 0.00164 & 0.00159 & 0.00154 & 0.00149 \\
\hline [tex]-2.8[/tex] & 0.00256 & 0.00248 & 0.00240 & 0.00233 & 0.00226 & 0.00219 & 0.00212 & 0.00205 \\
\hline [tex]-2.7[/tex] & 0.00347 & 0.00336 & 0.00326 & 0.00317 & 0.00307 & 0.00298 & 0.00289 & 0.00280 \\
\hline [tex]-2.6[/tex] & 0.00466 & 0.00453 & 0.00440 & 0.00427 & 0.00415 & 0.00402 & 0.00391 & 0.00379 \\
\hline [tex]-2.5[/tex] & 0.00621 & 0.00604 & 0.00587 & 0.00570 & 0.00554 & 0.00539 & 0.00523 & 0.00508 \\
\hline
\end{tabular}
\][/tex]


Sagot :

To determine how many students scored less than 96 on this standardized test, let's follow these steps systematically:

1. Determine the z-score:
- The z-score formula is given by:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
where [tex]\( x \)[/tex] is the score to find (96), [tex]\( \mu \)[/tex] is the mean score (156), and [tex]\( \sigma \)[/tex] is the standard deviation (23).

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

2. Find the cumulative probability associated with the z-score:
- Using the provided z-score table, we observe that our exact z-score (-2.6087) is not available in the table. However, we can determine its probability through interpolation using nearby values.

- From the table:
[tex]\[ \begin{aligned} \text{For } z = -2.6: & \quad \text{Probability} = 0.00403 \\ \text{For } z = -2.7: & \quad \text{Probability} = 0.00326 \\ \end{aligned} \][/tex]

- We need a value between [tex]\(-2.6\)[/tex] and [tex]\(-2.7\)[/tex]. Interpolation can be done as:
[tex]\[ \text{Interpolated probability for } z = -2.6087 \approx 0.003169 \][/tex]

3. Calculate the number of students corresponding to this cumulative probability:
- The cumulative probability tells us the proportion of students scoring less than a particular value.
- Multiply this probability by the total number of students (48,592):
[tex]\[ \text{Number of students} = 0.003169 \times 48,592 \approx 154 \][/tex]

So, approximately 154 students scored less than 96 on this standardized test.