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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 71 and 10, 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 71 and 10, respectively. If the scores are normally distributed, what percentage of students scored below 73 ?

Option: 1

57.93 \%


Option: 2

56.36 \%


Option: 3

69.33 \%


Option: 4

75.89 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 73 , we can use the properties of the normal distribution.

First, we need to calculate the Z-score for a score of 73. 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=(73-71) / 10 \\ & Z=2 / 10 \\ & Z=0.2 \end{aligned} }

Next, we need to find the cumulative probability associated with the Z-score of 0.2 . This represents the percentage of values below 73 .

Using a standard normal distribution table or a calculator, we can find that the cumulative probability associated with a Z-score of 0.2 is approximately 0.5793 .

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

\mathrm{ \begin{aligned} & \text { Percentage }=0.5793 \times 100 \\ & \text { Percentage } \approx 57.93 \% \end{aligned} }
Therefore, approximately 57.93 \% of students scored below 73 on the test.

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vishal kumar

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