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

Solution :

Given that ,

mean = = 7.37

standard deviation = = 0.75

a)

P(5 < x < 6) = P((5 - 7.37)/ 0.75) < (x - ) /  < (6 - 7.37) / 0.75) )

= P(-3.16 < z < -1.83)

= P(z < -1.83) - P(z < -3.16)

= 0.03360 - 0.00080 Using z- table.

= 0.03280

The probability that their hospital stay is from 5 to 6 days is 0.03280.

b)

P(x > 6) = 1 - P(x < 6)

= 1 - P((x - ) / < (6 - 7.37) / 0.75)

= 1 - P(z < -1.83)

= 1 - 0.03360 Using standard normal table.

= 0.96640

The probability that their hospital stay is greater than 6 days is 0.96640.