Question

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)

Answer 1

i) Check attached image.

ii) P(x < 2) = 0.0122

iii) Check attached image.

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

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!!!

Answer 2

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) ]\\\\=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]