Question

A study was made of seat belt use among children who were involved in car crashes that caused them to be hospitalized. It was found that children not wearing any restraints had hospital stays with a mean of 7.37 days and a standard deviation of 1.50 days with an approximately normal distribution.

(a) Find the probability that their hospital stay is from 5 to 6 days, rounded to five decimal places.

(b) Find the probability that their hospital stay is greater than 6 days, rounded to five decimal places.

Answer 1

Given that children not wearing any restraints had hospital stays with a mean of = 7.37 days and a standard deviation of = 1.50 days with an approximately normal distribution.

(a) So the probability that their hospital stay is from 5 to 6 days P(5<X<6) is calculated by finding the Z score at 5 and 6 as:

and at X = 6

So, P(5<X<6) is transposed to Z scores as P(-1.58<Z<-0.91) is calculated using excel formula for normal distribution which is =NORM.S.DIST(-0.91, TRUE)-NORM.S.DIST(-1.58, TRUE) thus P(5<X<6) computed as 0.12436.

(b) The probability that their hospital stay is greater than 6 days P(X>6) is computed by finding the Z score as:

so, P(X>6) = P(Z>-0.91) is computed using excel formula =1-NORM.S.DIST(-0.91, TRUE) which results in probability as 0.81859.