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bezimeni [28]
3 years ago
8

Hurry please in test rn!!

Mathematics
1 answer:
natulia [17]3 years ago
3 0
Answer y=2x+3

explanation
y=mx+b

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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:

<u>The complete question is:</u> 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 = <u><em>the heights of men in the​ 20-29 age group</em></u>

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 = <u>0.4013</u>

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 = <u>0.5586</u>.

(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 = <u>0.0401</u>

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
Are the statements true or false?
My name is Ann [436]

false false false false false

5 0
4 years ago
Two trains leave the station at the same time, one heading east and the other west. The eastbound train travels at 100 miles per
storchak [24]

Solving for the elapsed time

distance traveled = (speed of first train + speed of second train) * elapsed time

280 miles = (95 mph + 105 mph) * elapsed time

280 miles = 200 mph * elapsed time

elapsed time = 280 miles / 200 mph

elapsed time = 1.4 hours

elapsed time = 1 hour and 24 minutes

It will take 1 hour and 24 minutes for the two trains to be 280 miles apart.

7 0
3 years ago
Complete the equations.<br> look at photo
Jlenok [28]

Answer:

25.8

258.

2580.

Step-by-step explanation:

just move the decimal that many times

8 0
3 years ago
Read 2 more answers
What is the equation of a line parallel to y = x+2<br> and passes through (1, 2)?
kicyunya [14]

Answer: y - x = 1

Step-by-step explanation: Note : for two lines to be parallel, then they have the same slope.

i.e the slope is 1

Using y - y1 = m ( x-x1) to find the equation of the line

⇒ y - 2 = 1 (x - 1)

    y - 2 = x - 1

    y - x = -1 +2

    y - x = 1

it could also be written in slope intercept form as y = x + 1

7 0
3 years ago
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