Question

In: Statistics and Probability

The amount of time that people spend at Grover Hot Springs is normally distributed with a...

The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 80 minutes and a standard deviation of 14 minutes. Suppose one person at the hot springs is randomly chosen. Let X = the amount of time that person spent at Grover Hot Springs . Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(,)

b. Find the probability that a randomly selected person at the hot springs stays longer then 89 minutes.

c. The park service is considering offering a discount for the 4% of their patrons who spend the least time at the hot springs. What is the longest amount of time a patron can spend at the hot springs and still receive the discount?  minutes.

d. Find the Inter Quartile Range (IQR) for time spent at the hot springs.
Q1:  minutes
Q3:  minutes
IQR:  minutes

Solutions

Expert Solution

Solution :

Given that ,

mean = = 80

standard deviation = = 14

a) The distribution of x is normal X N ( 80, 14 )

b) P(x > 89) = 1 - p( x< 89)

=1- p P[(x - ) / < (89 - 80) / 14]

=1- P(z < 0.64)

Using z table,

= 1 - 0.7389

= 0.2611

c) Using standard normal table,

P(Z < z) = 4%

= P(Z < z) = 0.04  

= P(Z < -1.75) = 0.04

z = -1.75

Using z-score formula,

x = z * +

x = -1.75 * 14 + 80

x = 55.5 minutes.

d) Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.67 ) = 0.25

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 * 14 + 80

x = 70.62

First quartile =Q1 = 70.62 minutes

The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.67 ) = 0.75

z = 0.67

Using z-score formula,

x = z * +

x = 0.67 * 14 + 80

x = 89.38

Third quartile =Q3 = 89.38 minutes

IQR = Q3 - Q1

IQR = 89.38 - 70.62

IQR = 18.76 minutes


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