Question

A computerized driving test held ten sessions in a day. The overall passing rate of the driving test is 87.5%. In each session, there are 14 candidates. (a) Most likely, how many candidates would pass the test in a session? What is the probability of its happening? (b) What is the probability that more than 10 candidates in a session can pass the test? (c) What are the (i) expectation, (ii) variance, and (iii) standard deviation of number of candidates can pass the driving test in a day. (d) Suppose each candidate who fail the driving test will immediately join the tutorial class for the 2nd test. The charge of the tutorial class is $400. What are the (i) expectation and (ii) standard deviation of the daily income earned from the tutorial class?

Answer 1

A computerized driving test held ten sessions in a day.

The overall passing rate of the driving test is 87.5%.

Thus p = 0.875      ( Probability of success )

There are 14 candidates , thus n =14

(a) Most likely, how many candidates would pass the test in a session

There are 10 session a day .

Total number of candidates ,in each session = 14

Passing rate of the driving test is 87.5%.

Candidates who can pass the test in a session = 14 * 87.5 /100 = 12.25 12 to 13

Most likely 12 to 13 candidates would pass the test in a session

(b) What is the probability that more than 10 candidates in a session can pass the test

Let X be candidate , probability of passing driving test is 0.875

Thus X ~ B( n = 14 , p = 0.875   )

To find more than 10 candidates in a session can pass the test

Thus we need to find P(X > 10 )

Here P(X=x) =

Thus P(X > 10 ) = P( X = 11 ) + P( X = 12 ) + P( X = 13 ) + P( X = 14 )

                     = +

   + +

                         = 0.1636515 + 0.2863901 + 0.3084201 + 0.1542101

        P(X > 10 ) = 0.9126718

Hence , probability that more than 10 candidates in a session can pass the test is 0.9126718.

(c) What are the (i) expectation, (ii) variance, and (iii) standard deviation of number of candidates can pass the driving test in a day.

(i) expectation - Mean(X) = n*p

                                               = 14 * 0.875 = 12.25

Thus , expectation   = 12.25        { each session }

One every day there are 10 ten sessions

In each session, there are 14 candidates.

Number of candidates every day = 10 * 14 = 140

Thus number of expected candidates from total of 140 candidates that can pass exam in a day is 123

{ 12.25 * 140=122.5 123 }

(ii) variance, - Var(X) = np(1-p)

                                      = 14 * 0.875 * 0.125 = 1.53125

Thus , variance   = 1.53125    { each session }

           variance = (n*10)*p*q = 15.3125         { For each day }

(iii) standard deviation -

Thus , standard deviation = 1.237437 { for each session } as = 1.237437   

              standard deviation = 3.913119 { For each day } as = 3.913119

(d) Suppose each candidate who fail the driving test will immediately join the tutorial class for the 2nd test. The charge of the tutorial class is $400. What are the (i) expectation and (ii) standard deviation of the daily income earned from the tutorial class?

Probability that candidate fails the driving test is 1-0.875 = 0.125

Thus probability fails the driving test is 0.125

In each session, there are 14 candidates.

Driving test held ten sessions in a day

Thus total number of candidates = 14 * 10 = 140

Expected Number of candidates fail the driving test = 140 * 0.125 = 17.5 18

Candidate who fail the driving test will immediately join the tutorial class for the 2nd test.

The charge of the tutorial class is $400

Thus money earned from failed students = 400 * 17.5 = 7000

i.e suppose on firest dat 17.5 { not in real stituation take it as aprrox 18 } fails the test then total money earned from those failed candidates is $7000

Thus

(i) expectation of the daily income earned from the tutorial class is $7000

(ii) To find standard deviation of the daily income earned from the tutorial class

Note - variation for number of failed candidates will be same that to number of passing candidates

As    Failing candidates = 140 - Passing candidates

    var(Failing candidates) = var(140 - Passing candidates)

                                                = var(Passing candidates) = var(X)

    var(Failing candidates) = var(X)

We have obtaing var(X)= 15.3125         { For each day }

Thus var(Failing candidates) = 15.3125   { For each day }

Now The charge of the tutorial class is $400

There fore variation in chareges taken from failed students is

Var(Z) = var(400 * Failing candidates) = 4002 * Var(X) = 4002 * 15.3125 = 2450000

Now Standard deviation = = = 1565.248

Thus Standard deviation 1565.248.

Therefore , if each candidate who fail the driving test will immediately join the tutorial class for the 2nd test. and the charge of the tutorial class is $400 .

Then (i) expectation of the daily income earned from the tutorial class is $7000

and (ii) standard deviation of the daily income earned from the tutorial class is 1565.248.