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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 66 and 20, respectively. If the scores are normally distributed, what percentage

In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 66 and 20, respectively. If the scores are normally distributed, what percentage of students scored below 68?

 

Option: 1

64.32%


Option: 2

98.45%


Option: 3

72.11%


Option: 4

54.99%


Answers (1)

best_answer

To determine the percentage of students who scored below 68, we can use the properties of a normal distribution. Given that the mean (\mu) is 66 and the standard deviation (\sigma) is 20, we can calculate the z-score corresponding to a score of 68 and then find the percentage of scores below that z-score.

The z-score is calculated using the formula:

x is the individual score
\mu is the mean
\sigma is the standard deviation
In this case, we want to find the z-score for x=68:

\begin{aligned} & z=(68-66) / 20 \\ & z=0.1 \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 \mathrm{z}-score of 0.1 is approximately 54.99%.

Therefore, approximately 54.99% of students scored below 68.

 

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