Get Answers to all your Questions

header-bg qa

In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 65 and 8 , respectively. If the scores are normally distributed, what percentage of students scored below 70 ?

Option: 1

73.40 \%


Option: 2

78.14 \%


Option: 3

72.15 \%


Option: 4

86.65 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 70 , we need to calculate the z-score corresponding to that value and then find the area under the normal curve to the left of that z-score.

The z-score is calculated using the formula:

\mathrm{ z=(x-\mu) / \sigma }
where:
x is the value we want to find the percentage below (in this case, 70),
\mu is the mean (65), and
\sigma is the standard deviation (8).
Plugging in the values:

\mathrm{ \begin{aligned} & z=(70-65) / 8 \\ & z=5 / 8 \\ & z=0.625 \end{aligned} }
Now we need to find the area to the left of this z-score in the standard normal distribution table (also known as the z-table) or by using a statistical software. The area represents the percentage of students who scored below 70 .

Looking up the z-score of 0.625 in the z-table, we find that the area to the left is approximately 0.7340 .

So, approximately 73.40% of the students scored below 70 on the test.

 

Posted by

Ritika Kankaria

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE