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he amount of time people spend exercising in a given week follows a normal distribution with a mean of 3.8 hours per week and a standard deviation of 0.8 hours per week. i. Which of the following shows the shaded probability that a person picked at random exercises less than 2 hours per week? a. b. ii. What is the probability that a person picked at random exercises less than 2 hours per week? (round to 4 decimal places) iii. Which of the following shows the shaded probability that a person picked at random exercises between 2 and 4 hours per week? a. b. iv. What is the probability that a person picked at random exercises between 2 and 4 hours per week? (round to 4 decimal places)

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

Let X denote the amount of time spending exercise in a given week

Given that X normal (3.8, (0.8)²)

Thus we know that

$Z= (x-3.8)/(0.8) N(0,1)$

i)P [ amount of time less than two hour ]

= P[x < 2]

ii)

$P [x < 2]=P[(x-3.8)/(0.8) < (2-3.8)/(0.8) ]\n\n=P[z<-2.25]$

= P[z > 2.25] ∴ symmetric

= P[0 ≤ z ≤ ∞] - P[0 ≤ z ≤ 2.25]

= 0.5 - 0.48778

= 0.0122

iii)

P[2 < x < 4]

Answer 2

Answer:

i) Check attached image.

ii) P(x < 2) = 0.0122

iii) Check attached image.

iv) P(2 < x < 4) = 0.5865

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 3.8 hours per week

Standard deviation = σ = 0.8 hours per week

i) The probability that a person picked at random exercises less than 2 hours per week on a shaded graph?

P(x < 2)

First of, we normalize/standardize the 2 hours per week

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (2.0 - 3.8)/0.80 = -2.25

The probability that someone picked at random exercises less than 2 hours weekly is shown in the attached image to this question.

P(x < 2) = P(z < -2.25)

ii) To determine the probability that someone picked at random exercises less than 2 hours weekly numerically

P(x < 2) = P(z < -2.25)

We'll use data from the normal probability table for these probabilities

P(x < 2) = P(z < -2.25) = 0.01222 = 0.0122 to 4 d.p

iii) The probability that a person picked at random exercises between 2 and 4 hours per week on a shaded graph?

P(2 < x < 4)

We then normalize or standardize 2 hours and 4 hours.

For 2 hours weekly,

z = -2.25

For 4 hours weekly,

z = (x - μ)/σ = (4.0 - 3.8)/0.80 = 0.25

The probability that someone picked at random exercises between 2 and 4 hours weekly is shown in the attached image to this question.

P(2 < x < 4) = P(-2.25 < z < 0.25)

iv) To determine the probability that a person picked at random exercises between 2 and 4 hours per week numerically

P(2 < x < 4) = P(-2.25 < z < 0.25)

We'll use data from the normal probability table for these probabilities

P(2 < x < 4) = P(-2.25 < z < 0.25)

= P(z < 0.25) - P(z < -2.25)

= 0.59871 - 0.01222 = 0.58649 = 0.5865

Hope this Helps!!!

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Use the following four coordinates to determine which one has a distance of 5 units from point A (2, 6). (5 points) (5, 2) (2, 5) (6, 7) (7, 5)

Answers

A right triangle with legs measuring 3 units and 4 units has a hypotenuse measuring 5 units.

You need to find a point which has a difference in x and y of 3 and 4 or 4 and 3 from the point A(2, 6).

Look at (5, 2) and compare with A(2, 6).

Difference in x: 5 - 2 = 3

Difference in y: 6 - 2 = 4

Since the differences in x and y are 3 and 4, the hypotenuse will measure 5.

Answer: (5, 2)

Suppose that an experiment has five possible outcomes, which are denoted {1,2,3,4,5}. Let A be the event {1,3,4} and let B be the event {2,4,5}. (Notice that we did not say that the five outcomes are equally likely: the probability distributions could be anything.) For each of the following relations, tell whether it could possibly hold. If it could, give a numerical example using a probability distribution of your own choice: if it could not, explain why not (what rule is violated)a. P(A) = P(B)
b. P(A) = 2P(B)
c. P(A) = 1 - P(B)
d. P(A) + P(B) > 1
e. P(A) - P(B) < 0
f. P(A) - P(B) > 1

Answers

Answer:

a. P(A) = P(B)

c. P(A) = 1 - P(B)

a and c are true . The rest are false.

Step-by-step explanation:

Two events A and B are said to be equally likely when one event is as likely to occur as the other. In other words each event should occur in equal number in repeated trials. For example when a fair coin is tossed the head is likely to appear as the tail, and the proportion of times each side is expected to appear is 1/2.

So when the events A= {1,3,4} B = {2,4,5} are equally likely then suppose their probability is 1/2.

a. P(A) = P(B) True

1/2= 1/2

b. P(A) = 2P(B) False

1/2 is not equal to 1

c. P(A) = 1 - P(B) True

1/2= 1-1/2= 1/2

d. P(A) + P(B) > 1 False

1/2 + 1/2 is not greater than 1

e. P(A) - P(B) < 0 False

1/2-1/2= 0 is not less than 0

f. P(A) - P(B) > 1 False

1/2-1/2= 0 is not greater than 1

Final answer:

The relationships between the probabilities are evaluated and explained.

Explanation:

a. P(A) = P(B) could possibly hold if P(A) = 1/3 and P(B) = 1/3.

b. P(A) = 2P(B) could not hold, as probabilities cannot exceed 1.

c. P(A) = 1 - P(B) could possibly hold if P(A) = 2/3 and P(B) = 1/3.

d. P(A) + P(B) > 1 could possibly hold if P(A) = 1/3 and P(B) = 1/2.

e. P(A) - P(B) < 0 could not hold, as the difference between probabilities cannot be negative.

f. P(A) - P(B) > 1 could not hold, as the difference between probabilities cannot exceed 1.

Learn more about Probability here:

brainly.com/question/22962752

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A company produces two products, A and B. The sales volume for A is at least 80% of the total sales of both A and B. However, the company cannot sell more than 110 units of A per day. Both products use one raw material, of which the maximum daily availability is 300 lb. The usage rates of the raw material are 2 lb per unit of A, and 4 lb per unit of B The profit units for A and B are $40 and $90, respectively. Determine the optimal product mix for the company

Answers

Answer:A=100 , b=25

Step-by-step explanation:

Let sales of A be x and sales of B be y

Thus $x\geq 0.8(x+y)$

$x\geq 4y$

Also maximum A available is $110\geq x$

$300\geq 2x+4y$

$150\geq x+2y$

We have find the optimal solution for

z=40x+90y

Optimal solution points

(100,25) z $=40* 100+90* 25=6250$

(110,20) z $=40* 110+90* 20=6200$

(110,0) z $=40* 110+90* 0=4400$

Thus for A=100 and B=25 Optimal solution is obtained

Final answer:

The optimal product mix problem involves maximizing profit given certain constraints. The constraints can be expressed in terms of inequalities which can be solved using linear programming techniques such as the corner point theorem or the simplex method.

Explanation:

The subject of this problem is to determine the optimal product mix of two products, A and B, produced by a company. This is guided by several constraints including sales volumes, maximum output, raw material availability, and profit units.

From the problem, we have two constraints. Firstly, sales of A must be at least 80% of the total sales of A and B, and no more than 110 units of A can be sold per day. Secondly, the company cannot use more than 300 lbs of the raw material per day with usage rates of 2 lbs per unit of A and 4 lbs per unit of B.

Let the quantity of A and B sold per day be x and y respectively. The profit is given by the expression 40x + 90y. We need to maximize this expression based on the constraints. The constraints can be expressed as follows:

x ≥ 0.8(x + y), this is the sales volume constraint.
x ≤ 110, this is the maximum sales constraint.
2x + 4y ≤ 300, this constraint arises from the raw material availability.

These constraints form a linear programming problem. By plotting these inequalities on a graph and finding the feasible region, we can use the corner point theorem or simplex method to find the optimal solution.