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Andreyy89
3 years ago
12

Height of Dutch Men. Males in the Netherlands are the tallest, on average, in the world with an average height of 183 centimeter

s (cm) (BBC News website). Assume that the height of men in the Netherlands is normally distributed with a mean of 183 cm and standard deviation of 10.5 cm.a.What is the probability that a Dutch male is shorter than 175 cm?b.What is the probability that a Dutch male is taller than 195 cm?c.What is the probability that a Dutch male is between 173 and 193 cm?d.Out of a random sample of 1000 Dutch men, how many would we expect to be taller than 190 cm?
Mathematics
1 answer:
trapecia [35]3 years ago
3 0

Answer:

a) P(X

P(z

b) P(X>195) =P(Z>\frac{195-183}{10.5})=P(Z>1.143)

P(z>1.143)=1-P(Z

c) P(173

P(-0.953

P(-0.953

d) P(\bar X >190)=P(Z>\frac{190-183}{\frac{10.5}{\sqrt{1000}}}=21.08)

And using a calculator, excel ir the normal standard table we have that:

P(Z>21.08)=1-P(Z

So for this case we expect anyone with a heigth higher than 190 cm in a random sample of 1000.

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 heights of a population, and for this case we know the distribution for X is given by:

X \sim N(183,10.5)  

Where \mu=183 and \sigma=10.5

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 with the normal standard table or with excel:

P(z

Part b

P(X>195)

P(X>195) =P(Z>\frac{195-183}{10.5})=P(Z>1.143)

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

P(z>1.143)=1-P(Z

Part c

P(173

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(173And we can find this probability on this way:

P(-0.953

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

P(-0.953Part d

Since the distributiion for X is normal then the distribution for the sample mean is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we eant this probability:

P(\bar X >190)=P(Z>\frac{190-183}{\frac{10.5}{\sqrt{1000}}}=21.08)

And using a calculator, excel ir the normal standard table we have that:

P(Z>21.08)=1-P(Z

So for this case we expect anyone with a heigth higher than 190 cm in a random sample of 1000.

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The amounts invested in stocks, bonds, and CDs were $45,000, $80,000, and $10,000, respectively.

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Alternatively, you can reduce the augmented matrix for this problem to row-echelon form using any of several calculators or on-line sites. That matrix is ...

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