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 59 and 13 , 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 59 and 13 , respectively. If the scores are normally distributed, what percentage of students scored below 63 ?

Option: 1

55.63 \%


Option: 2

85.69 \%


Option: 3

77.36 \%


Option: 4

61.94 \%


Answers (1)

best_answer

To determine the percentage of students who scored below 63 , we can use the properties of the normal distribution.

First, we need to calculate the Z-score for a score of 63 . 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=(63-59) / 13 \\ & Z=4 / 13 \\ & Z \approx 0.3077 \end{aligned} }

Next, we need to find the cumulative probability associated with the Z-score of 0.3077 . This represents the percentage of values below 63 .

Using a standard normal distribution table or a calculator, we can find that the cumulative probability associated with a Z-score of 0.3077 is approximately 0.6194 .

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

\mathrm{ \begin{aligned} & \text { Percentage }=0.6194 \times 100 \\ & \text { Percentage } \approx 61.94 \% \end{aligned} }
Therefore, approximately 61.94 \% of students scored below 63 on the test.

Posted by

Ritika Jonwal

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

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question