Question

Question 19 A study was made of seat belt use among children who were involved in car Type numbers in the boxes. crashes that caused them to be hospitalized. It was found that children not Part 5 points wearing any restraints had hospital stays with a mean of 7.37 days and a Part 5 points standard deviation of 2.60 days with an approximately normal distribution. 10 points (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

Here it is given that distribution is normal with mean=7.37 and standard deviation=2.60

a. We need to find

As distribution is normal we can convert x to z

,5 – 7.37 PC 2.60 P(5 < 3 < 6) -0.53) = 0.11665 6 – 7.37 2.60 P(-0.91 <2<

b. Here we need to find $P(x>6)=P(z>\frac{6-7.37}{2.60})=P(z>-0.53)=0.70194$

ん

So is the answer