Question

A) The total distances covered by GIMPA Undergraduate students to and from their work
places every month is normally distributed with a mean 105 km and variance of 225 km2
.
a. What percentage of the population of students covered less than 90 km in a month in other
words what is the probability of finding a student who covers less than 90 km in a month?
(1 mark)
b. The percentage of students who cover above 140 km every month (1 mark)
c. The percentage of students who cover between 100 and 120 km every month (1 mark)
d. What is the least distance of the top 0.1 decile category of the distances covered by students
in a month? (2 mark)
B) A health clinic found that in a sample of 200 women, the mean and the standard deviation
of their masses are 85.5kg and 10.5kg, respectively. Also, 115 women were found to be over-
weight.
a. Find point estimates of the mean mass and proportion of over-weight in women population.
(1 mark)
b. Construct a 95% confidence interval for population mean mass. ( 2 marks)
c. Construct a 95% confidence for proportion of over-weight in women population.

Answer 1

Que 1)The total distances covered by GIMPA Undergraduate students to and from their work
places every month is normally distributed with a mean 105 km and variance of 225 km2

Solutions→

Let x→ denote the distance

X~N(105,225)

Mean(mu)=

225

Sigma= 15

Formula→z= (x-mu)/sigma

a)what is the probability of finding a student who covers less than 90 km in a month.

Prob(X<90)

Prob((X-mu)/sigma<(90-105)/15)

=Prob(Z<-1)

=0.1587

(b)The percentage of students who cover above 140 km every month

Prob(x>140)

= Prob((z>(140-105)/15))= prob(z>2.33)=0.0099

Thus ans→0.99 percentage≈1percentage

C)The percentage of students who cover between 100 and 120 km every month

Prob(100<x<120)

= Prob(x<120)-prob(x<100)

=0.8412-0.9693

=0.4719

47.19%(ans)

d)What is the least distance of the top 0.1 decile category of the distances covered by students
in a month

Let the least distance is x'

Prob(X>x')= 0.1

Prob(Z>(x'-mu)/sigma)= 0.1

From z table→ (x'-105)/15=1.282

This imply. x'=124.233(ans)

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