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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 70 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 70 and 8 , respectively. If the scores are normally distributed, what percentage of students scored below 80 ?

Option: 1

52.33
 


Option: 2

96.45
 


Option: 3

72.22
 


Option: 4

89.44


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{Z=Z \text {-score }}

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

Plugging in the values: 

\mathrm{\begin{aligned} & Z=(80-70) / 8 \\ & Z=10 / 8 \\ & Z=1.25 \end{aligned}}

Next, we need to find the cumulative probability associated with the Z-score of 1.25 . 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 1.25 is approximately 0.8944 .
To convert this probability to a percentage, we multiply it by 100 :

\mathrm{ \begin{aligned} & \text { Percentage }=0.8944 \times 100 \\ & \text { Percentage } \approx 89.44 \% \end{aligned} }
Therefore, approximately 89.44 \% of students scored below 80 on the test.

Posted by

Sumit Saini

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