For this exercise please refer to the downloaded Excel Workbook (See Excel Workbook Instructions) from the JB Learning website with your access code. Please upload your answers as an excel file with properly labeled answers. At the end of the chapter 5 workbook you will find Practice problems. Please answer in excel only under practice problems 5.4 questions: 1, 2, 3, 4, 5. If help is needed, tutorials are found in the workbook with relevant problems.Note: These pictures will give you an idea where it is in the workbook. You are responsible to use the workbook.#1. Total cholesterol in children aged 1015 years is assumed to follow a normal distribution with a mean of191 and a standard deviation of 22.4.what proportion of children 1015 years of age hfave total cholesterol between 180-190?What proportion of children 10-15 years would be classified as hyperlipidemia(assume that hyperlipidemiais defined as a total cholesterol level over 200)?What is the 90th percentile of total cholesterol?Solution:We have μ = 191 and σ = 22.4a) The respective Z-score with X = 180 and 190 isZ = (X -μ)/σ = (180 – 191)/22.4 = 0.49Z = (X -μ)/σ = (190 – 191)/22.4 = 0.04Therefore, using Z-tablesP [0.49

Question

For this exercise please refer to the downloaded Excel Workbook (See Excel Workbook Instructions) from the JB Learning website with your access code. Please upload your answers as an excel file with properly labeled answers. At the end of the chapter 5 workbook you will find Practice problems. Please answer in excel only under practice problems 5.4 questions: 1, 2, 3, 4, 5. If help is needed, tutorials are found in the workbook with relevant problems.

Note: These pictures will give you an idea where it is in the workbook. You are responsible to use the workbook.

#1. Total cholesterol in children aged 1015 years is assumed to follow a normal distribution with a mean of

191 and a standard deviation of 22.4.

what proportion of children 1015 years of age hfave total cholesterol between 180-190?
What proportion of children 10-15 years would be classified as hyperlipidemia(assume that hyperlipidemia

is defined as a total cholesterol level over 200)?

What is the 90th percentile of total cholesterol?

Solution:

We have

μ = 191 and σ = 22.4

a) The respective Z-score with X = 180 and 190 is

Z = (X -μ)/σ

= (180 – 191)/22.4

= 0.49

Z = (X -μ)/σ

= (190 – 191)/22.4

= 0.04

Therefore, using Z-tables

P [0.49 < Z < 0.04] =0.5160 – 0.3121

= 0.2039

= 20.39%

b) The respective Zscore with X = 200 is

Z = (X -μ)/σ

= (200 – 191)/22.4

= 0.40

Therefore, using Z-tables

P [Z > 0.40] = 1 – 0.6554

= 0.3446 or 34.46%

c) The respective Z-score with p < 0.90 is 1.282

We have

Z = (X-μ)/σ

1.282 = (X – 191)/22.4

1.282*22.4 + 191 = X

X = 219.72

Among Coffee drinkers, men drank a mean of 3.2 cups per day with a standard deviation of 0.8 cups .

Assume the number of coffee drinks per day follows a normal distribution.

what proportion drinks 2 cups per day or more?
What proportion drank no more than 4 cups per day?
if the top 5% of coffee drinkers are considered heavy coffee drinkers , what is the minimum number of

cups consumed by a heavy coffee drinker? (Hint: find the 95th percentile)

Solution:

We have

μ= 3.2 and σ= 0.8

a) The respective Z-score with X = 2 is

Z = (X -μ)/σ

= (2 – 3.2)/0.8

= 1.5

Therefore, using Z-tables

P [Z ≥1.5] = 0.9332 or 93.32%

b) The respective Z-score with X = 4 is

Z = (X -μ)/σ

= (2 – 3.2)/0.8

= 1.5

Therefore, using Z-tables

P [Z ≥1.5] = 0.9332 or 93.32%

c) The respective Zscore with p < 0.96 is 1.645

We have

Z = (X -μ)/σ

1.645 = (X – 191)/22.4

1.645*22.4 + 191 = X

X = 227.85

A study is conducted to assess the imopact of caffeine consumption, smoking, alcohol consumption, andphysical activity on the risk of cardiovascular disease, Suppose that 40% of participants consume caffeine and smoke . If 8 participants are evaluated, what is the probability that:
exactly half of them consume caffeine and smoke?
at most 6 consume caffeine and smoke?

Solution:

We have n = 8 and p = 0.40. Using binomial distribution,

P [X = x] =n C x p x (1 – p) n-x

a) P [X = 4] = ⁸C₄ (0.40) ⁴(1 – 0.40) ^8-4

= ⁸C ₄(0.40) ⁴(0.60)⁴

= 0.2322

b) P [X ≤ 6] = 1 – P [X > 6]

=1 – [P (X = 7) + P (X = 8)]

=1 – (0.0079 + 0.0006)

= 0.9915

A recent study of cardiovascular risk factors reports that 30% of adults meet the criteria for hypertention. If

15 adults are assessed, what is the probability that :

Exactly 5 meet the criteria for hypertension
none meet the criteria for hypertension
less than or equal to 7 meet the criteria for hypertension

Solution:

We have n = 15 and p = 0.30. Using binomial distribution,

P [X = x] =ⁿC x p^x (1 – p) ^n-x

a) P [X = 5] = ¹⁵C ₅(0.30) 5 (1 – 0.30) ^15-5

= ¹⁵ C 5(0.30) ⁵ (0.70) ¹⁰

= 0.2061

b) P [X = 0] = 15 C 0 (0.30) 0(1 – 0.30) 15-0

= ¹⁵ C ₀ (0.30) ⁰(0.70) ¹⁵

= 0.0047

c) P [X ≤7] = P [X = 0] + P [X = 1] + P [X = 2] + P [X = 3] + P [X = 4] + P [X = 5] + P [X = 6] + P [X = 7]

= 0.0047 + 0.0305 + 0.0916 + 0.1700 + 0.2186 + 0.2061 + 0.1472 + 0.0811

= 0.9499

= 0.95

Diastolic Blood pressure are assumed to follow a normal distribution with a mean of 85 and a standar

deviation of 12.

WHAT proportion of people have diastolic blood pressures less than 90?
what proportion have diastolic blood pressures between 80 and 90?
If someone has a diastolic blood pressure of 100, what percentile does this represent.?

Solution:

We have

μ = 85 and σ = 12