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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?

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