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inessss [21]
3 years ago
13

In an all boys school, the heights of the student body are normally distributed with a mean of 70 inches and a standard deviatio

n of 5 inches. Using the empirical rule, what percentage of the boys are between 65 and 75 inches tall?
Mathematics
2 answers:
Zepler [3.9K]3 years ago
6 0

Answer:

68%

Step-by-step explanation:

Ket [755]3 years ago
3 0

Answer:

68,3 %

Step-by-step explanation:

Empirical rule establishes:

In a normal distribution with mean μ and standard deviation σ we have a  relation between values of the distribution and intervals describes as follows:

( μ ± σ )  contains  68.3 % of all values of the distribution

( μ ± 2σ )  contains 95,4 %       and

( μ ± 3σ )  contains 99.7 %

In our particular case we have that

( μ ± σ )   ⇒    (  70  ±  5 )   ⇒   ( 65 , 75 )

( μ ± 2σ ) ⇒    ( 70   ±  10 ) ⇒   ( 60 , 80 )     and

( μ ± 3σ ) ⇒    ( 70  ±  15  )  ⇒   ( 55 , 85

As we can clearly see 68,3 % of values fall between values going from 65 up to 75 inches

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The amount of time that people spend at Grover Hot Springs Spa is normally distributed with a mean of 63 minutes and a standard
PtichkaEL [24]

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

3 0
3 years ago
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