Mathematics, 17.03.2020 02:01 ahjd2020

In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 69.5inches and a standard deviation of 3.0inches. A study participant is randomly selected. Complete parts (a) through (d) below.

a. Find the probability that a study participant has a height that is less than 66 inches.
b. Find the probability that a study participant has a height that is between 66 and 71 inches.
c. Find the probability that a study participant has a height that is more than 71 inches.
d. Identify any unusual events. Explain your reasoning.

Answers: 3

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Answers

Answer from: Valduarte07

a) P(X

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

P(z

b) P(66

And we can find this probability with thi difference:

P(-1.167

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

P(-1.167

c) $P(X71)=P(\frac{X-\mu}{\sigma}\frac{71-\mu}{\sigma})=P(Z\frac{71-69.5}{3.0})=P(z0.5)$

And we can find this probability using the complement rule and the normal standard table:

P(z0.5)=1-P(z

d) $\mu - 2*\sigma = 69.5 -2*3=63.5$

$\mu + 2*\sigma = 69.5 + 2*3=75.5$

If a value is lower than 63.5 or higher than 75.5 we can consider it as unusual

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:

$X \sim N(69.5,3.0)$

Where $\mu=69.5$ and $\sigma=3.0$

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:

P(z

Part b

P(66

And we can find this probability with thi difference:

P(-1.167

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

P(-1.167

Part c

$P(X71)=P(\frac{X-\mu}{\sigma}\frac{71-\mu}{\sigma})=P(Z\frac{71-69.5}{3.0})=P(z0.5)$

And we can find this probability using the complement rule and the normal standard table:

P(z0.5)=1-P(z

Part d

For this case we can consider as unusual events values above/ below two deviations within the mean:

$\mu - 2*\sigma = 69.5 -2*3=63.5$

$\mu + 2*\sigma = 69.5 + 2*3=75.5$

If a value is lower than 63.5 or higher than 75.5 we can consider it as unusual

Answer from: Quest

Answer: d. 48 step-by-step explanation: i used a calculator so i am 100% sure these are right mark as brainliest?

Answer from: Quest

4 candies

step-by-step explanation:

7 x 28 = 196

200 - 196 = 4 candies remaining.

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