Homework Help

An examination is marked out of 100. It is taken bby a large number of candidates. The...

user profile pic

bookenghua | Student, Undergraduate | eNoter

Posted May 6, 2013 at 5:06 AM via web

dislike 1 like

An examination is marked out of 100. It is taken bby a large number of candidates. The mean for all candidates is 72.1 and the standard deviation is 15.2. Give a reason why a normal distribution with this mean and standard distribution, would not give a good approximation to the distribution of marks.

Thanks.

1 Answer | Add Yours

user profile pic

pramodpandey | College Teacher | Valedictorian

Posted May 6, 2013 at 5:54 AM (Answer #1)

dislike 1 like

Standard normal random variable

`Z=(X-mu)/sigma`

`We`  have given `mu=72.1 and sigma=15.2`

Thus

`Z=(100-72.1)/15.2=1.84`

`Z=(0-72.1)/15.2=-4.74`

Thus

`P(0<=X<=100)=.967`

Thus

`P(X<0)+P(X>100)=1-.967=.033`

ie. 3.3% which is ,non negligible probabilty 3.3 % significant

Thus normal is non fit for this marks distribution. 

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes