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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 74 and 11 , respectively. If the scores are normally distributed, what percentag

In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 74 and 11 , respectively. If the scores are normally distributed, what percentage of students scored below 80 ?

Option: 1

36.15 \%


Option: 2

66.14 \%


Option: 3

70.68 \%


Option: 4

86.96 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 80 , we can use the properties of the normal distribution.

First, we need to calculate the Z-score for a score of 80 . The Z-score represents the number of standard deviations a particular value is from the mean.

The formula to calculate the Z-score is:

\mathrm{ Z=(X-\mu) / \sigma }
Where:

\mathrm{ \begin{aligned} & Z=Z-\text { score } \\ & X=\text { Score } \\ & \mu=\text { Mean } \\ & \sigma=\text { Standard deviation } \end{aligned} }

Plugging in the values:

\mathrm{ \begin{aligned} & Z=(80-74) / 11 \\ & Z=6 / 11 \\ & Z \approx 0.5455 \end{aligned} }

Next, we need to find the cumulative probability associated with the Z-score of 0.5455 . This represents the percentage of values below 80 .

Using a standard normal distribution table or a calculator, we can find that the cumulative probability associated with a Z-score of 0.5455 is approximately 0.7068 .

To convert this probability to a percentage, we multiply it by 100 :

\mathrm{ \begin{aligned} & \text { Percentage }=0.7068 \times 100 \\ & \text { Percentage } \approx 70.68 \% \end{aligned} }
Therefore, approximately  70.68 \% of students scored below 80 on the test.

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