PSR Mock test Solution

Attempted on best effort basis. Need cross verifictions

Question 1

Virat Kohli’s scores in the last ten innings are as follows: 78, 94, 134, 56, 2, 67, 89, 152, 26, 42 Based on the above data, compute the expected value and standard deviation of Virat
Kohli’s batting score. (8 Marks)

Solution

Expected Value, (X) = µ = Σx / n , cosidering that the probability of each value in the set is same.

virat_scores = c(78,    94, 134,    56, 2,  67, 89, 152,    26, 42)
mean(virat_scores)
## [1] 74

Standard Deviation , σ = √V(X) v(X) = Σ(x-E(x))^2 / n

var(virat_scores)
## [1] 2134.444

So Standard Deviations is,

sd(virat_scores)
## [1] 46.20005

Question 2

The amount of regular unleaded gasoline purchased every week at a gas station near UCLA follows the normal distribution with mean 50000 gallons and standard deviation 10000
gallons. The starting supply of gasoline is 74000 gallons, and there is a scheduled weekly
delivery of 47000 gallons.
a. Find the probability that, after 11 weeks, the supply of gasoline will be below 20000
gallons.
b. b. How much should the weekly delivery be so that after 11 weeks the probability that the supply is below 20000 gallons is only 0.5%? (12 Marks)

Solution for a

Source : http://www.stat.ucla.edu/~nchristo/statistics100A/stat100a_clt.pdf

Facts :

µ = 50000, σ = 10000 , n = 11 , the purchase follows normal distribution

What we need to find out ? By the end of 11 week, what is the probability that supply is less than 20000.

For supply probability to be less that 20000, we need to understand what is the corresponding purchase probability after 11 weeks to deplete the supply below the levels of 20000. This is crucial to understand.

So,

Total Supply for 11 weeks = Initial Supply + 11 * recurring supply every week.

total_supply <- 74000 + 11 * 47000
total_supply
## [1] 591000

The Minimum Purchase to be made during 11 weeks to deplete the stock below 20000 is ,

total_min_purchase <- total_supply - 20000
total_min_purchase
## [1] 571000

So we need to find the probability of purchase greater than 571000

Now as purchase follow normal distribution, the cumulative probability could be applied using z-score table. Lets get corresponding Z score:

z = (total_min_purchase - mean of 11 week purchases) / Standard Deviation over 11 weeks =

(total_min_purchase - 11*50000)/(sqrt(11)*10000)
## [1] 0.6331738

P (Z > 0.6331) = 1 - P(Z < 0.6331)

From Standard Normal Table, P(Z < 0.6331) = 0.7357

Therefore , **The Probability of purchase of quantity of gasoline over 11 weeks which amounts to deplete of supply less than 20000 is ,

1- 0.7357
## [1] 0.2643

Solution for b

Let X be the weekly supply at which the the probability of supply below 20000 by end of 11 weeks is 0.5%. That is p = 0.005

that is , total purchase greater than 74000 + 11 * X - 20000 to be at probability 0.005

The z value for probability of 0.005 is 2.575 . Converting this into actual numbers,

2.575 = ((74000 + 11 * X - 20000) - 11 * 50000) / (sqrt(11) * 10000)

Question 3

If the average number of claims handled daily by an insurance company is 5, what
proportions of days have less than 3 claims? What is the probability that there will be 4
claims in exactly 3 of the next 5 days? Assume that the number of claims on different days is
independent. (8 Marks)

Solution

µ = 5

As only average claims info is provided, the best option is to apply Poisson Distribution here. So, lamba = 5

what proportions of days have less than 3 claims?

P(X < 3) = P(0) + P(1) + P(2)

ppois(2, lambda = 5)
## [1] 0.124652

What proportion of 4 claims in 3 of next 5 days

In 3 days out of 5, 4 claims to be handled daily. The probability of 4 claims per day could be arrived by poisson distribution. Now that probability could be applied in binimial distribution to see what is the chances of 4 claims for 3 days out of 5.

dbinom(3,5,dpois(4, lambda = 5))
## [1] 0.03672864

Question 4

An architect is designing a doorway for a public building to be used by people whose heights are normally distributed, with mean 1 meter 75 centimeter, and standard deviation 7.5
centimeter. How long can the doorway be so that no more than 1 % of the people bump
their heads? (8 Marks)

Solution

µ = 1 metre 75 centimeter = 175 σ = 7.5

And Heights of people follow normal distribution

Length of doorway so that not more than 1% people bump their head is nothing but doorway length should be higher than 99% people height.

The cumulative probability of 0.99 in terms of z score is 2.33

The corresponding value of height X is ,

(2.33 * 7.5) + 175
## [1] 192.475

Or using formula for higher precision,

qnorm(0.99, 175, 7.5)
## [1] 192.4476

Hence the height of door should be 192.475 centimeter for 99% of people to pass through without bumping