Question

A) The amount of time it takes students to complete a quiz is uniformly distributed between 35 to 65 minutes. One student is selected at random. Find the following events: a. The probability density function (pdf) of amount of time taken for the quiz. b. The probability that the student requires more than 50 minutes to complete the quiz.(1 mark) c. The probability that completed quiz time is between 45 and 55 minutes. (1 mark) d. The probability that the student requires at least 40 minutes to complete the quiz.(1 mark) e. The probability that the student completes the quiz exactly 60 minutes. (1 mark) f. Compute the expected time and the coefficient of variation to complete the quiz. B) Using 92% confidence, how large a sample should be taken to obtain a margin of error for the estimation of a population proportion of 0.045? Assume that the data are not available for developing the planning value for ?.

Answer 1

here for uniform distribution parameter a =35 and b=65

a) e probability density function (pdf) :

f(x)=1/(b-a) =1/30 for 35<x<65

b)

P(X>50)=-P(X<50)=1-(50-35)/(65-35)=0.5

c)

P(45<X<55)=(55-45)/(65-35)=0.3333

d)

P(X>40)=-P(X<40)=1-(40-35)/(65-35)=0.8333

e)

P(X=60) =0 (since point probability on a continuous distribution is 0)

f)expected time:

μ=(a+b)/2 =

50

standard deviation σ=(b-a)/√12=

8.6603

coefficient of variation =( σ/µ)*100 =17.32 %

B)

here margin of error E =	0.045
for92% CI crtiical Z          =	1.751
estimated prop.=p=	0.500
sample size n= p*(1-p)*(z/E)2=	379