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

Answer:

a) 0.03310

b) 0.96610

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =7.37 days

Standard Deviation, σ =  0.75 days

We are given that the distribution of  hospital stays is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( hospital stay is from 5 to 6 days)

$P(5 \leq x \leq 6) = P(\displaystyle\frac{5 - 7.37}{0.75} \leq z \leq \displaystyle\frac{6-7.37}{0.75}) = P(-3.16 \leq z \leq -1.827)\\\\= P(z \leq -1.827) - P(z < -3.16)\\= 0.034 - 0.001 = 0.03310 = 3.31\%$

$P(5 \leq x \leq 6) = 3.3\%$

b) P(hospital stay is greater than 6 days)

P(x > 6)

$P( x > 6) = P( z > \displaystyle\frac{6 - 7.37}{0.75}) = P(z > -1.827)$

$= 1 - P(z \leq -1.827)$

Calculation the value from standard normal z table, we have,  

$P(x > 6) = 1 - 0.0339 = 0.96610 = 96.61\%$