Question

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.

Answer 1

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

Answer 2

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

Answer 3

My answer for c.)

0.4772

Answer 4

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.

Answer 5

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

Answer 6

part b.)

p should be 0.1915

so 19 students

Answer 7

no not 19 students