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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 72 and 8 , 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 72 and 8 , respectively. If the scores are normally distributed, what percentage of students scored below 75 ?

Option: 1

96.14 \%


Option: 2

64.71 \%


Option: 3

96.85 \%


Option: 4

54.36 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 75 , we can use the properties of the normal distribution.

First, we need to calculate the Z-score for a score of 75 . 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=(75-72) / 8 \\ & Z=3 / 8 \\ & Z=0.375 \end{aligned} }

Next, we need to find the cumulative probability associated with the Z-score of 0.375 . This represents the percentage of values below 75 .

Using a standard normal distribution table or a calculator, we can find that the cumulative probability associated with a Z-score of 0.375 is approximately 0.6471 .

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

\mathrm{ \begin{aligned} & \text { Percentage }=0.6471 \times 100 \\ & \text { Percentage } \approx 64.71 \% \end{aligned} }
Therefore, approximately 64.71 \% of students scored below 75 on the test

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manish painkra

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