Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 53 and 16 , 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 53 and 16 , respectively. If the scores are normally distributed, what percentage of students scored below 58 ?

Option: 1

61.84 \%


Option: 2

46.65 \%


Option: 3

78.98 \%


Option: 4

96.36 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 58 , 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, 58),

\mu is the mean (53), and

\sigma is the standard deviation (16).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(58-53) / 16 \\ & z=5 / 16 \\ & z \approx 0.3125 \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 58 .

Looking up the z-score of 0.3125 in the z-table, we find that the area to the left is approximately 0.6184 .

So, approximately 61.84% of the students scored below 58 on the test.

 

Posted by

jitender.kumar

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

BTech Admission 2025 Live: JEE Main correction window closes at 11:50 pm; VITEEE, SRMJEE, AEEE dates out
anam-khan

JEE Main 2025 application form correction begins at jeemain.nta.nic.in; editable fields
anam-khan

JEE Main 2025 application form correction window opens today; apply at jeemain.nta.nic.in

Latest Question