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

Answer:

a)$X \sim N(63,18)$  

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

b) $P(X>90)=P(\frac{X-\mu}{\sigma}>\frac{90-\mu}{\sigma})=P(Z>\frac{90-63}{18})=P(Z>1.5)$

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) $P(X<45)=P(\frac{X-\mu}{\sigma}<\frac{45-\mu}{\sigma})=P(Z<\frac{45-63}{18})=P(Z<-1)$

$P(Z<-1)=0.159$

d) $P(60<X<90)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{90-\mu}{\sigma})=P(\frac{60-63}{18}<Z<\frac{90-63}{18})=P(-0.167<z<1.5)$

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

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

e)  IQR = 75.13-50.87=24.26

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

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

$P(X>90)=P(\frac{X-\mu}{\sigma}>\frac{90-\mu}{\sigma})=P(Z>\frac{90-63}{18})=P(Z>1.5)$

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:

$z=\frac{x-\mu}{\sigma}$

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

$P(X<45)=P(\frac{X-\mu}{\sigma}<\frac{45-\mu}{\sigma})=P(Z<\frac{45-63}{18})=P(Z<-1)$

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:

$P(60<X<90)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{90-\mu}{\sigma})=P(\frac{60-63}{18}<Z<\frac{90-63}{18})=P(-0.167<z<1.5)$

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:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25$  

$P(z<\frac{a-\mu}{\sigma})=0.25$

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

$z=-0.674<\frac{a-63}{18}$

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

$z=0.674<\frac{a-63}{18}$

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