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In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 76 and 18, respectively. If the scores are normally distributed, what percentage of students scored below 80?

 

Option: 1

63.14%


Option: 2

96.12%


Option: 3

72.45%


Option: 4

58.24%


Answers (1)

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To determine the percentage of students who scored below 80, we can use the properties of a normal distribution. Given that the mean (\mu) is 76 and the standard deviation (\sigma) is 18, we can calculate the z-score corresponding to a score of 80 and then find the percentage of scores below that z-score.

The z-score is calculated using the formula:

\mathrm{z=(x-\mu) / \sigma}

Where:
\mathrm{x} is the individual score
\mathrm{\mu}  is the mean
\mathrm{\sigma} is the standard deviation

In this case, we want to find the z-score for  \mathrm{x= 80}:

\begin{aligned} & z=(80-76) / 18 \\ & z=0.2222 \end{aligned}

Once we have the z-score, we can look up the corresponding percentage in the standard normal distribution table or use a calculator that provides this functionality. From the table, we find that the percentage of scores below a z-score of 0.2222 is approximately 58.24%.

Therefore, approximately 58.24% of students scored below 80.

 

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