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Vsevolod [243]
2 years ago
11

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.
Mathematics
1 answer:
PtichkaEL [24]2 years ago
3 0

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

c) P(X

P(Z

d) P(60

P(-0.167

P(-0.167

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

Part c

We are interested on this probability

P(X

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

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

P(Z

Part d

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

P(60

And we can find this probability with this difference:

P(-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

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   (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  

P(z

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

z=-0.674

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

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

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