Question

your help is appreciated :)

The Fullerton Arboretum is a wonderful place to visit. It is located on the CSUF campus. The amount of time spent at the Arboretum is normally distributed with a mean of 61 minutes and a standard deviation of 15 minutes. Suppose one person at the Arboretum is randomly selected. Let X = the amount of time that person spent at the Fullerton Arboretum. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X - NG b. Find the probability that a randomly selected person at the Fullerton Arboretum stays longer then 57 minutes. c. The school is considering offering a discount for the 7% of their patrons who spend the least time at the Arboretum. What is the longest amount of time a patron can spend at the Arboretum and still receive the discount? minutes. d. Find the Inter Quartile Range (IQR) for time spent at the Arboretum. Q1: minutes Q3: minutes IQR: minutes

Answer 1

a) X ~ N(61, (15)²)

b) P(X > 57)

X- 57 - P o

$= P\left(Z > \frac{57 - 61}{15} \right )$

= P(Z > -0.27)

= 1 - P(Z < -0.27)

= 1 - 0.3936

= 0.6064

c) P(X < x) = 0.07

$or, P\left(\frac{X - \mu}{\sigma} < \frac{x - \mu}{\sigma} \right ) = 0.07$

$or, P\left(Z < \frac{x - 61}{15} \right ) = 0.07$

C - 61 or, -1.48 15

d) P(X < x) = 0.25

$or, P\left(\frac{X - \mu}{\sigma} < \frac{x - \mu}{\sigma} \right ) = 0.25$

$or, P\left(Z < \frac{x - 61}{15} \right ) = 0.25$

$or, \frac{x - 61}{15}=-0.67$

Q1 = 50.95

P(X < x) = 0.75

$or, P\left(\frac{X - \mu}{\sigma} < \frac{x - \mu}{\sigma} \right ) = 0.75$

$or, P\left(Z < \frac{x - 61}{15} \right ) = 0.75$

$or, \frac{x - 61}{15}=0.67$

Q3 = 71.05

IQR = Q3 - Q1 = 71.05 - 50.95 = 20.1