Question

The amount of time that people spend at Grover Hot Springs Spa is normally distributed with a mean of 63 minutes and a standard deviation of 18 minutes. Answer the following, rounding probabilities to 4 decimal places. Let X= the amount of time that a person spends at Grover Hot Springs Spa a. The distribution is X ~ ( , ) b. What's the probability that a randomly chosen person spends more than 90 minutes at the Spa ?c. What's the probability that a randomly person spends less than 45 minutes at the Spa?d. What's the probability that a randomly person spends between 60 and 90 minutes at the Spa?e. Find the IQR. minutes Round answers to 2 decimals.

Answer 1

a)
`X \sim N(63,18)`

Where
`\mu=63` and
`\sigma=18`

b)
`P(X>90)=P((X-\mu)/(\sigma)>(90-\mu)/(\sigma))=P(Z>(90-63)/(18))=P(Z>1.5)`

And we can find this probability using the complement rule and with the normal standard table or excel:

`P(Z>1.5)=1-P(Z<1.5)=1-0.933=0.067`

c)
`P(X<45)=P((X-\mu)/(\sigma)<(45-\mu)/(\sigma))=P(Z<(45-63)/(18))=P(Z<-1)`

`P(Z<-1)=0.159`

d)
`P(60<X<90)=P((60-\mu)/(\sigma)<(X-\mu)/(\sigma)<(90-\mu)/(\sigma))=P((60-63)/(18)<Z<(90-63)/(18))=P(-0.167<z<1.5)`

`P(-0.167<z<1.5)=P(z<1.5)-P(z<-0.167)`

`P(-0.167<z<1.5)=P(z<1.5)-P(z<-0.167)=0.933-0.434=0.499`

e) IQR = 75.13-50.87=24.26

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the amount of time that people spend at Grover Hot Springs of a population, and for this case we know the distribution for X is given by:

`X \sim N(63,18)`

Where
`\mu=63` and
`\sigma=18`

Part b

We are interested on this probability

`P(X>90)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X>90)=P((X-\mu)/(\sigma)>(90-\mu)/(\sigma))=P(Z>(90-63)/(18))=P(Z>1.5)`

And we can find this probability using the complement rule and with the normal standard table or excel:

`P(Z>1.5)=1-P(Z<1.5)=1-0.933=0.067`

Part c

We are interested on this probability

`P(X<45)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<45)=P((X-\mu)/(\sigma)<(45-\mu)/(\sigma))=P(Z<(45-63)/(18))=P(Z<-1)`

And we can find this probability using the normal standard table or excel:

`P(Z<-1)=0.159`

Part d

If we apply this formula to our probability we got this:

`P(60<X<90)=P((60-\mu)/(\sigma)<(X-\mu)/(\sigma)<(90-\mu)/(\sigma))=P((60-63)/(18)<Z<(90-63)/(18))=P(-0.167<z<1.5)`

And we can find this probability with this difference:

`P(-0.167<z<1.5)=P(z<1.5)-P(z<-0.167)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-0.167<z<1.5)=P(z<1.5)-P(z<-0.167)=0.933-0.434=0.499`

Part e

Q1

For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.75` (a)

`P(X<a)=0.25` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.25`

`P(z<(a-\mu)/(\sigma))=0.25`

But we know which value of z satisfy the previous equation so then we can do this:

`z=-0.674<(a-63)/(18)`

And if we solve for a we got

`a=63 -0.674*18=50.87`

So the value of height that separates the bottom 25% of data from the top 75% is 50.87.

Q3

Since the distribution is symmetrical we repaeat the procedure for Q1 but now with z= 0.674

`z=0.674<(a-63)/(18)`

And if we solve for a we got

`a=63 +0.674*18=75.13`

And then the IQR = 75.13-50.87=24.26