Question

2. A college graduate wants to buy a car but is concerned about the rising gas prices in the city of Tema. It is known the mean gas price is 1.55 CAD/L with a standard deviation of 0.2 CAD/L. The graduate wil (a) Assuming that the gas prices in the city of Tema follows a Normal distribution, find the following i. A gas station is selected at random from the city of Tema. What is the probability that the price ii. Eight gas stations are randomly selected in the city of Tema. What is the probability that the buy the car if the gas prices are below 1.40 CAD/L at the end of the year. probabilities: of gas wil be less than 1.25 CAD/L? [2 marks] average gas price among the sampled gas stations is between 1.45 CAD/L to 1.60 CAD/L?2 marks iii. If forty gas stations in the city of Tema are randomly selected, what is the probability that the average gas price among the sampled gas stations is between 1.48 CAD/L to 1.58 CAD/L? 12 (b) If the gas price in the city of Tema DOES NOT follow a normal distribution, will the probabilities (c) It is known that there is a 60% chance of the college graduate buying the car. Do you think the marks computed in (a) be still accurate? Briefly explain your answer for each. 3 marks prices of gas in the city of Tema follow the Normal model? Justify why or why not. [2 marks

Answer 1

Define random variable X : Gas price.

a) Assume that the gas price follows Normal (mean = 1.55, std deviation = 0.2).

i) Here we have to find the probability the price of gas is less than 1.25 CAD/L. That means

P(X<1.25) = P((X-mean/SD) <(1.25-1.55)/0.2)

                 = P(Z< -1.5)

                = 0.0668                     (From statistical table of areas to the left of Z)

Probability that the price of gas is less than 1.25 CAD/L is 0.0668

ii) Sample size : n= 8

Here we have to find probability that the average gas price among sampled gas stations is between 1.45 CAD/L and 1.60 CAD/L.

Mean of sample(X bar) follows Normal (mean=1.55, Std deviation = 0.2/sqrt(n)=0.07)

P(1.45< xbar<1.60) = P((1.45-1.55)/0.07 <Z<(1.60-1.55)/0.07)

                                 =P( -1.43< Z <0.71)

                                =P(Z<0.71) - P(Z<-1.43)        (Round to two decimal places)

                               =0.7611 - 0.0764

                              = 0.6847

Probability hat the average gas price among sampled gas stations is between 1.45 CAD/L and 1.60 CAD/L is 0.6847.

iii) Here, n = 40

We have to find probability that the average price among sampled gas stations is between 1.48 CAD/L and 1.58 CAD/L.

X bar follows normal (mean = 1.55, SD = 0.2/sqrt(40) = 0.032)

P(1.48<X bar < 1.58) = P((1.48-1.55)/0.032 < X bar < (1.58-1.55)/0.032)

                                   =P(-2.19 < Z < 0.94)

                                   =P(Z<0.94)-P(Z< -2.19)      (Round to two decimal places)

                                  =0.9738-0.0143

                                = 0.9595

Probability that the average price among sampled gas stations is between 1.48 CAD/L and 1.58 CAD/L is 0.9595.

b) Probability answers for first part will be change if we did not assume normal distribution.

In part ii) we have use the Central limit theorem (CLT). But CLT is applicable only when n is very large (n>30).

In part iii) we have used CLT which is appropriate because here n is greater than 30. Hence answer will remain same.

Statement of Central limit theorem :When sample size (n) is large the sampling distribution of sample mean is approximately normal with standard deviation = SD/sqrt(n).

c) P( college graduate buying the car) = 0.6

The graduate will buy the car if the gas prices are below 1.40CAD/L                (Given information)

That means here two outcomes are possible: will buy car(Success) and will not buy car (failure).

That means we can use geometric distribution with p=0.6.