MATHEMATICS COLLEGE

The sizes of houses in Kenton County are normally distributed with a mean of 1346square feet with a standard deviation of 191 square feet. For a randomly selected
house in Kenton County, what is the probability the house size is:
a. over 1371 square feet?
O Z=
o probability =
b. under 1296 square feet?
O Z=
o probability =
c. between 773 and 1637 square feet?
o zl =
o Z2 =
o probability =
Note: Z-scores should be rounded to 2 decimal places & probabilities should be
rounded to 4 decimal places.
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Points possible: 8
This is attempt 1 of 3.

Answers

Answer 1

Answer:

Answer:

(a) The probability that the house size is over 1371 square feet is 0.4483.

(b) The probability that the house size is under 1296 square feet is 0.3974.

(c) The probability that the house size is between 773 and 1637 square feet is 0.9344.

Step-by-step explanation:

We are given that the sizes of houses in Kenton County are normally distributed with a mean of 1346 square feet with a standard deviation of 191 square feet.

Let X = the sizes of houses in Kenton County

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

Z = $(X-\mu)/(\sigma)$ ~ N(0,1)

where, $\mu$ = mean size of houses = 1346 square feet

$\sigma$ = standard deviation = 191 square feet

(a) The probability that the house size is over 1371 square feet is given by = P(X > 1371 square feet)

P(X > 1371) = P( $(X-\mu)/(\sigma)$ > $(1371-1346)/(191)$ ) = P(Z > 0.13) = 1 - P(Z $\leq$ 0.13)

= 1 - 0.5517 = 0.4483

The above probability is calculated by looking at the value of x = 0.13 in the z table which has an area of 0.5517.

(b) The probability that the house size is under 1296 square feet is given by = P(X < 1296 square feet)

P(X < 1296) = P( $(X-\mu)/(\sigma)$ < $(1296-1346)/(191)$ ) = P(Z < -0.26) = 1 - P(Z $\leq$ 0.26)

= 1 - 0.6026 = 0.3974

The above probability is calculated by looking at the value of x = 0.26 in the z table which has an area of 0.6026.

(c) The probability that the house size is between 773 and 1637 square feet is given by = P(773 square feet < X < 1637 square feet)

P(773 < X < 1637) = P(X < 1637) - P(X $\leq$ 773)

P(X < 1637) = P( $(X-\mu)/(\sigma)$ < $(1637-1346)/(191)$ ) = P(Z < 1.52) = 0.9357

P(X $\leq$ 773) = P( $(X-\mu)/(\sigma)$ $\leq$ $(773-1346)/(191)$ ) = P(Z $\leq$ -3) = 1 - P(Z $\leq$ 3)

= 1 - 0.9987 = 0.0013

The above probabilities are calculated by looking at the value of x = 1.52 and x = 3 in the z table which has an area of 0.9357 and 0.9987 respectively.

Therefore, P(773 square feet < X < 1637 square feet) = 0.9357 - 0.0013 = 0.9344.

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Step-by-step explanation:

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Answers

Answer with Step-by-step explanation:

Given

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2. A faucet for a 30-gallon bathtub fills at a rate of 3 gallons per minute.If the drain lets water out at 1.5 gallons per minute, how long would it
take for the tub to overflow if both the faucet and drain were open at the
same time?

Answers

Answer:

Step-by-step explanation:

Knowing that it is a 30-gallon bathtub.

And it fills at a rate of 3 gallons per minute.

But also drains 1.5 gallons per minute.

So this is for the first minute.

3 - 1.5 = 1.5

So there will be 1.5 gallons of water in the tub after the first minute.

Now do,

30 ÷ 1.5 = 20

It will take 20 full minutes for the bathtub to fill up.

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Answers

Answer:

What's the question?

Step-by-step explanation:

Put it DB and I'll gladly answer!

A circle is divided into 6 equal parts. What is the angle measure of 1 part?

Answers

The angle measure of one part will be 60.

360/6 = 60

That is because the circle it divided into 6 equal parts and the whole circle is 360.

Hope this helps :)

A company that manufactures toothpaste is studying five different package designs. Assuming that one design is just as likely to be selected by a consumer as any other design, what selection probability would you assign to each of the package designs? We would assign a probability of to the design 1 outcome, to design 2, to design 3, to design 4, and to design 5. In an actual experiment, 100 consumers were asked to pick the design they preferred. The following data were obtained. Design Number of Times Preferred 1 10 2 5 3 30 4 40 5 15 Do the data confirm the belief that one design is just as likely to be selected as another? Explain. Yes, the sum of the assigned probabilities is 1. No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely. Yes, the average of the assigned probabilities is 0.20. No, a probability of about 0.50 would be assigned using the relative frequency method if selection is equally likely.

Answers

Answer:

Correct option: "No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely."

Step-by-step explanation:

The assumption made is that all the 5 different packages are equally likely, i.e. the probability of selecting a package is $(1)/(5)=0.20$ .

The probability distribution is shown below.

According to the probability distribution:

The probability of a person preferring design 1 is,

$P(X=1)=0.10$

The probability of a person preferring design 2 is,

$P(X=2)=0.05$

The probability of a person preferring design 3 is,

$P(X=3)=0.30$

The probability of a person preferring design 4 is,

$P(X=4)=0.40$

The probability of a person preferring design 1 is,

$P(X=5)=0.15$

So it can be seen that the probability of preferring any of the 5 designs are not same.

Thus, the designs are not equally likely.

The correct option is "No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely."

The selection Probability determined using the relative frequency method do not match the assigned probabilities, suggesting that the data do not confirm the belief that one design is as likely to be selected as another.

The given data can be used to calculate the relative frequencies of each package design selected by the consumers.

To determine the selection probabilities using the relative frequency method, divide the number of times a design was preferred by the total number of consumers.

For example, for design 1, the selection probability would be 10/100 = 0.1.

Similarly, for design 2, the selection probability would be 5/100 = 0.05.

The selection probabilities for designs 3, 4, and 5 would be 0.3, 0.4, and 0.15 respectively.

Comparing these probabilities to the assigned probabilities, it can be observed that the assigned probabilities do not match the observed relative frequencies, indicating that the data do not confirm the belief that one design is just as likely to be selected as another.

Learn more about Probability here:

brainly.com/question/22962752

#SPJ3

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