PLEASE HELP ME FOR EXTRA POINTS AND BRAINLIEST ANWSER

n a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of inches and a stand

kotykmax [81]

Answer:

(a) The probability that a study participant has a height that is less than 67 inches is 0.4013.

(b) The probability that a study participant has a height that is between 67 and 71 inches is 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is 0.0401.

(d) The event in part (c) is an unusual event.

Step-by-step explanation:

The complete question is: In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches. 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 67 inches. The probability that the study participant selected at random is less than inches tall is nothing. (Round to four decimal places as needed.) (b) Find the probability that a study participant has a height that is between 67 and 71 inches. The probability that the study participant selected at random is between and inches tall is nothing. (Round to four decimal places as needed.) (c) Find the probability that a study participant has a height that is more than 71 inches. The probability that the study participant selected at random is more than inches tall is nothing. (Round to four decimal places as needed.) (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.

We are given that the heights in the 20-29 age group were normally distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches.

Let X = the heights of men in the 20-29 age group

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean height = 67.5 inches

$\sigma$ = standard deviation = 2 inches

So, X ~ Normal( $\mu=67.5, \sigma^{2}=2^{2}$ )

(a) The probability that a study participant has a height that is less than 67 inches is given by = P(X < 67 inches)

P(X < 67 inches) = P( $\frac{X-\mu}{\sigma}$ < $\frac{67-67.5}{2}$ ) = P(Z < -0.25) = 1 - P(Z $\leq$ 0.25)

= 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 0.25 in the z table which has an area of 0.5987.

(b) The probability that a study participant has a height that is between 67 and 71 inches is given by = P(67 inches < X < 71 inches)

P(67 inches < X < 71 inches) = P(X < 71 inches) - P(X $\leq$ 67 inches)

P(X < 71 inches) = P( $\frac{X-\mu}{\sigma}$ < $\frac{71-67.5}{2}$ ) = P(Z < 1.75) = 0.9599

P(X $\leq$ 67 inches) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{67-67.5}{2}$ ) = P(Z $\leq$ -0.25) = 1 - P(Z < 0.25)

= 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 1.75 and x = 0.25 in the z table which has an area of 0.9599 and 0.5987 respectively.

Therefore, P(67 inches < X < 71 inches) = 0.9599 - 0.4013 = 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is given by = P(X > 71 inches)

P(X > 71 inches) = P( $\frac{X-\mu}{\sigma}$ > $\frac{71-67.5}{2}$ ) = P(Z > 1.75) = 1 - P(Z $\leq$ 1.75)

= 1 - 0.9599 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.9599.

(d) The event in part (c) is an unusual event because the probability that a study participant has a height that is more than 71 inches is less than 0.05.

7 0

3 years ago

A bag contains 5 black marbles, 9 white marbles, and 2 red marbles. One marble is to be chosen at random.

bulgar [2K]

2/16 because when you add all the marbles, you get 16 in total and there are 2 red marbles so there is a 2/16 chance of choosing one of the red marbles. If you want it simplified, it would be 1/8.

3 0

3 years ago

A single tree produces about 2.6x102 pounds of oxygen each year. the amazon rainforest has about 3.9 X 10 [ ] pounds per day ent

balu736 [363]

Answer:

The answer to your question is 2.8 x 10¹¹

Step-by-step explanation:

Data

1 tree produces 2.6 x 10² pounds of oxygen/year

number of trees = 3.9 x 10¹¹

pounds of oxygen per day = ?

Process

1.- Divide the pounds of oxygen by 365, to get the pounds of oxygen per day.

2.6 x 10² / 365 = 0.712

2.- Multiply the number of trees by the pounds of oxygen per day

3.9 x 10¹¹ x 0.712 = 2.8 x 10¹¹ pounds of oxygen

4 0

3 years ago

Last Sunday 1.575 people visited the amusement park % of the visitors were adults , 16% were teenagers and 28 were children ages

jasenka [17]

Answer:

882 , 252 , 441

Step-by-step explanation:

There are some mistakes in question and the corrected one is written below:

Q. Last Sunday 1,575 people visited the amusement park. 56% of the visitors were adults, 16% were teenagers, and 28% were children ages 12 and under. Find the number of adults, teenagers, and children that visited the park.

Given:

Total number of people visited amusement park = 1,575

Percent of adult people = 56%

Percent of teenagers = 16%

Percent of children = 28%

Now, we have to find number of adults, teenagers, and children that visited the park last Sunday.

Solution:

Number of adults = $56\%\ of \ 1575=\frac{56}{100}\times1575= \frac{88200}{100} =882\ adults$

Number of teenagers = $16\%\ of\ 1575=\frac{16}{100} \times1575=\frac{25200}{100} =252\ teenagers$

Number of children = $28\%\ of\ 1575=\frac{28}{100} \times1575=\frac{44100}{100} =441\ children$

Therefore, the number of adults are 882 , teenagers are 252 , and children are 441 that visited the park last Sunday.

6 0

3 years ago

I will give a brainliest if your answer are correct

Andrew [12]

Answer:

huhh what this is question is too hard

4 0

3 years ago

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