Question

The time taken (in minutes) by all students in a class to complete a test follows a normal distribution of 57 with variance of 34. a) A student is randomly chosen from the class.

Find the probability that

i) he/she takes more than 52 minutes to complete the test. (4 marks)

ii) he/she takes between 41 to 76 minutes to complete the test. (4 marks)

b) 5 students are chosen from the class at random, find the probability that at most 3 of them takes less than 61 minutes to complete the test. (7 marks)

c) 12 students are chosen from the class at random, find the probability at all of them takes between 43 to 64 minutes to complete the test. (5 marks) 

Answer 1

Solution:

Let X be the random variable denoting the time taken (in minutes) by all students in a class to complete a test.

$X\sim N(\mu,\sigma^2)$

with $\mu=57,\sigma^2=34$

(a) (i):

$z=\dfrac{X-\mu}{\sigma}$

Now, $P(X>52)=P(z>\dfrac{52-57}{\sqrt{34}})=P(z>-0.86)$

$=1-P(z\le-0.86)=1-0.19489=0.80511$

(ii):$P(41<X<76)=P(X<76)-P(X<41)$

$=P(z<\dfrac{76-57}{\sqrt{34}})-P(z<\dfrac{41-57}{\sqrt{34}})$

$=P(z<3.26)-P(z<-2.74) \\=0.99944-0.00307 \\=0.99637$

(b):

$P(X<61)=P(z<\dfrac{61-57}{\sqrt{34}})=P(z<0.69)=0.75490$

Now, assume a random variable $Y\sim Bin(n,p)$

Here, $n=5,p=0.7549,q=0.2451$

Then, $P(Y\le3)=1-P(Y>3)=1-[P(Y=4)+P(Y=5)]$

$=1-[^5C_4(0.7549)^4(0.2451)^1+^5C_5(0.7549)^5(0.2451)^0] \\=1-[5(0.7549)^4(0.2451)+(0.7549)^5] \\=0.6431$

(c):

$P(43<X<64)=P(X<64)-P(X<43)$

$=P(z<\dfrac{64-57}{\sqrt{34}})-P(z<\dfrac{43-57}{\sqrt{34}}) \\=P(z<1.2)-P(z<-2.4) \\=0.88493-0.0082 \\=0.87673$

Now, assume a random variable $R\sim Bin(n,p)$

Here $n=12,p=0.87676,q=0.12327$

Then, $P(R=12)=^{12}C_{12}(0.87676)^{12}(0.12327)^1=0.2063$