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The scores of the students on a standardized test are normally distributed, with a mean of 500 and a standard deviation of 110. What is the probability that a randomly selected student has a score between 350 and 550? Use the portion of the standard normal table below to help answer the question.

\begin{tabular}{|c|c|}
\hline
[tex]$z$[/tex] & Probability \\
\hline
0.00 & 0.5000 \\
\hline
0.25 & 0.5987 \\
\hline
0.35 & 0.6368 \\
\hline
0.45 & 0.6736 \\
\hline
1.00 & 0.8413 \\
\hline
1.26 & 0.8961 \\
\hline
1.35 & 0.9115 \\
\hline
1.36 & 0.9131 \\
\hline
\end{tabular}

A. [tex]$9\%$[/tex]

B. [tex]$24\%$[/tex]

C. [tex]$59\%$[/tex]


Sagot :

To find the probability that a randomly selected student has a score between 350 and 550 on a standardized test normally distributed with a mean of 500 and a standard deviation of 110, follow these steps:

1. Calculate the z-scores:
- The z-score represents the number of standard deviations a value is away from the mean.
- The formula for calculating the z-score is [tex]\( z = \frac{(X - \mu)}{\sigma} \)[/tex] where [tex]\( \mu \)[/tex] is the mean and [tex]\( \sigma \)[/tex] is the standard deviation.

For the lower score (350):
[tex]\[ z_{\text{lower}} = \frac{(350 - 500)}{110} = \frac{-150}{110} \approx -1.364 \][/tex]

For the upper score (550):
[tex]\[ z_{\text{upper}} = \frac{(550 - 500)}{110} = \frac{50}{110} \approx 0.455 \][/tex]

2. Use the z-table to find the corresponding probabilities:
- The z-table provides the cumulative probability for each z-score from the mean to the value.
- Approximate the given z-scores using the closest z-values from the table.

For [tex]\( z_{\text{lower}} \approx -1.364 \)[/tex], the closest given z-score is 0.00:
[tex]\[ P(Z < 0.00) = 0.5000 \][/tex]

For [tex]\( z_{\text{upper}} \approx 0.455 \)[/tex], the closest given z-score is 0.45:
[tex]\[ P(Z < 0.45) = 0.6736 \][/tex]

3. Calculate the probability between the z-scores:
- To find the probability that the score falls between the two z-scores, subtract the cumulative probability of the lower z-score from the cumulative probability of the upper z-score.

[tex]\[ P(350 < X < 550) = P(Z < 0.45) - P(Z < 0.00) \][/tex]

Substitute the values from the z-table:

[tex]\[ P(350 < X < 550) = 0.6736 - 0.5000 = 0.1736 \][/tex]

Therefore, the probability that a randomly selected student scores between 350 and 550 on this standardized test is approximately [tex]\( 0.1736 \)[/tex], or 17.36%.
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