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 0.75 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, Mean (u) = 7.37 days Standard Deviation (O) = 0.75 days The data of hospital stays is approximately a normal distribution. Let x denote the number of days a child stays at hospital. Also, z =* (a) Probability that their hospital stay is from 5 to 6 days = P[5<x<6] = P[x< 6] – P[x< ! P[z< -1.83] - P[z<-3.16] = 0.0336 -0.0008 = 0.03280 (b) Probability that their hospital stay is greater than 6 days = P[x >6] = P [2> 607;31) = P[z > -1.83] = 1 – P[z < -1.83] = 1 – 0.0336 = 0.96640