The marks of 100 students in an examination are normally distributed with mean of 25 marks and a standard deviation of 10 marks.
(a) Given that the pass mark is 20, estimate the number of students who passed the examination.
(b) If 5.59% of the students obtain a distinction by scoring x marks or more, estimate the value of x.
(c) If a sample of 20 students is selected, find the probability that the mean of the sample will be less than 27 marks.
My answer for a.)
P(X>20) = P[(x'-u)/(sd/sqrt(n)) > (20- 25)/(10/(sqrt100)
p(Z>-5) = 0.1915
1.0 - 0.1915 = 0.8085
0.8085 * 100 = 80.85 ~ 81students passed the test
My answer for b.)
5.59% = 0.0559
0.0559 ~ 0.14(frm Stand/Normal Table)
0.14 = (x-25)/(10/sqrt(100))
= 25.14
x = 25.14 marks
My answer for c.)
0.4772
a) Z = (score-mean)/SD
Z = (20-25)/10 (do not use SEm = SD/√n)
Z = -.5
p = .6915
p*n = ?
b) Again, don't use SEm, which is used for distribution of means rather than raw scores.
In the Z table, .0559 in the smaller portion gives a Z score of 1.59.
c) For this one, use SEm.
So taking note of the above corrections.
**Correction part a.)**
p = 0.6915
p * n = 0.6915 * 100
= 69.15 approx = 69
= 69 students passed the test
**Correction part b.)**
1.59 = (X-25)/10
= 40.9 marks
**Corrections part c.)
= (27-25)/10
z = 0.2; Z-table(0.0793)
= 0.0793
part b.)
p should be 0.1915
so 19 students