Question

The waiting time until a courier is received is followed by an exponential distribution with an expected waiting time of 5 days. Find the probability that the mean waiting time for the delivery time of 50 couriers exceeds 6 days

b. There are 200 questions to an FRCS exam. Suppose the probability of having through question correct is 0.6, no matter other questions. If the cutoff for a grade B is 80%, what is the likelihood of a "B"

Answer 1

(a) Here The waiting time until a courier is received is followed by an exponential distribution with an expected waiting time of 5 days.

so here x ~ EXPONETIAL (mean = 5 days, std. devition = sqrt(5))

as we take a sample of 50 couriers. so average waiting time for these 50 days will roughly follow normal distrbution with mean = 5 days and standard error = 5/sqrt(50) = 0.7071

~ NORMAL(5 ; 0.7071)

so here we have to find

P( > 6 days) =1 - P( < 6 days) =1 - NORMDIST(< 6 days; 5; 0.7071)

z = (6 - 5)/0.7071 = 1.4142

P( > 6 days) =1 - P( < 6 days) = 1- P(z < 1.4142) = 1 - NORMSDIST(1.4142) = 1- 0.9213 = 0.0786

(b) There are 200 questions to an FRCS exam. Suppose the probability of having through question correct is 0.6, no matter other questions.If the cutoff for a grade B is 80%.

so here expected percentage to be get = 060%

standard error of percentage to be get = sqrt(0.6 * 0.4/200) = 0.0346

P(Likelihood to get B) = P(p^ > 0.80 ; p = 0.6; se = 0.0346)

z = (0.80 - 0.6)/0.0346 = 5.7735

so here

P(Likelihood to get B) = P(Z > 5.7735) = 0.0000

very less probability