Question

(a) Susan tries to exercise at ”Pure Fit” Gym each day of the week, except on the weekends
(Saturdays and Sundays). Susan is able to exercise, on average, on 75% of the weekdays
(Monday to Friday).
i. Find the expected value and the standard deviation of the number of days she
exercises in a given week. [2 marks]
ii. Given that Susan exercises on Monday, find the probability that she will exercise at
least 3 days in the rest of the week. [3 marks]
iii. Find the probability that in a period of four weeks, Susan exercises 3 or less days in
only two of the four weeks

Answer 1

(a)

i)

Probability that Susan will exercise on a weekday : p = 75/100 =0.75

Number of days in a weekday : n =5

X : Number of days she exercises in weekdays

X follows Binomial distribution with n=5 and p =0.75

expected value of Binomial distribution = np

Standard deviation of the binomial distribution =

expected value of the number of days she exercises in a given week =np =5 x 0.75 = 3.75

Standard deviation of the number of days she exercises in a given week =

ii.

Given that Susan exercises on Monday, find the probability that she will exercise at
least 3 days in the rest of the week.

Given that Susan exercises on Monday, number of weekdays left in the rest of the week = 5-1 =4

Y : Number of days she exercises in the rest of week, given that Susan exercises on Monday

Y follows Binomial distribution with n=4 and p =0.75

Probability mass function of Y given by in the rest of week, given that Susan exercises on Monday

Probability that she exercise 'r' days in the rest of week, given that Susan exercises on Monday

Given that Susan exercises on Monday, probability that she will exercise at least 3 days in the rest of the week

= P(Y3) = P(Y=3)+P(Y=4)

P(Y3) = P(Y=3)+P(Y=4) = 0.421875+0.31640625=0.73828125

Given that Susan exercises on Monday, probability that she will exercise at least 3 days in the rest of the week = 0.73828125

iii.Find the probability that in a period of four weeks, Susan exercises 3 or less days in
only two of the four weeks

------

Probability that Susan will exercise on a weekday : p = 75/100 =0.75

Number of days in a weekday : n =5

X : Number of days she exercises in weekdays

X follows Binomial distribution with n=5 and p =0.75

p1 = Probability that Susan exercises 3 or less days in week : = P(X3)

P(X3) = 1 - [P(X=4)+ P(X=5)]

Probability that she exercise 'r' days in weekday

P(X3) = 1 - [P(X=4)+ P(X=5)] = 1 -[0.3955+0.2373] = 1-0.6328=0.3672

p1 = Probability that Susan exercises 3 or less days in week : = P(X3)

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p1 = Probability that Susan exercises 3 or less days in week : = 0.3672

n1 = 4 weeks

Z : number of weeks she exercises 3 or less days in 4 weeks

Z follow binomial distribution with n1 = 4 and p1 = 0.3672

Probability that Susan exercises 3 or less days in 'r' weeks of the four weeks is given by

Probability that in a period of four weeks, Susan exercises 3 or less days in only two of the four weeks = P(Z=2)

Probability that in a period of four weeks, Susan exercises 3 or less days in only two of the four weeks = 0.3240