The mean cholesterol levels of women age 45-59 in Ghana, Nigeria, and Seychelles is 5.1 mmol/l and the standard deviation is 1.0 mmol/l. Assume that cholesterol levels are normally distributed and a random sample of 21 women are selected. h) Find the probability that the sample mean cholesterol level of the 21 randomly selected women age 45-59 in Ghana, Nigeria, and Seychelles is more than 5.54 mmol/l.Round final answer to 4 decimal places.DO NOT use the rounded standard deviation from part e in this computation.Use the EXACT value of the standard deviation with the square root.i) Is a mean cholesterol level of 5.54 mmol/l unusually high for 21 randomly selected women age 45-59 inGhana, Nigeria, and Seychelles?Select an answerj) If you found a mean cholesterol level for a sample of 21 women age 45-59 in Ghana, Nigeria, andSeychelles as high as 5.54 mmol/l, what might you conclude?Select an answer d) Find the mean of the sampling distribution of the sample mean.Put the numeric value in the first box and the correct units in the second box.e) Find the standard deviation of the sampling distribution of the sample mean.Put the numeric value rounded to two decimal places in the first box and the correct units in the secondbox.f) What is the shape of the sampling distribution of the sample mean?Select an answerWhy? Check all that apply:population is normalo is unknownIn is less than 30population is not normalIn is at least 30o is knowng) Find the probability that the sample mean cholesterol level of the 21 randomly selected women age 45-59 in Ghana, Nigeria, and Seychelles is less than 4.7 mmol/l.Round final answer to 4 decimal places.DO NOT use the rounded standard deviation from part e in this computation.Use the EXACT value of the standard deviation with the square root.

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The mean cholesterol levels of women age 45-59 in Ghana, Nigeria, and Seychelles is 5.1 mmol/l and the standard deviation is 1.0 mmol/l. Assume that cholesterol levels are normally distributed and a random sample of 21 women are selected.

h) Find the probability that the sample mean cholesterol level of the 21 randomly selected women age 45-
59 in Ghana, Nigeria, and Seychelles is more than 5.54 mmol/l.
Round final answer to 4 decimal places.
DO NOT use the rounded standard deviation from part e in this computation.
Use the EXACT value of the standard deviation with the square root.
i) Is a mean cholesterol level of 5.54 mmol/l unusually high for 21 randomly selected women age 45-59 in
Ghana, Nigeria, and Seychelles?
Select an answer
j) If you found a mean cholesterol level for a sample of 21 women age 45-59 in Ghana, Nigeria, and
Seychelles as high as 5.54 mmol/l, what might you conclude?
Select an answer

d) Find the mean of the sampling distribution of the sample mean.
Put the numeric value in the first box and the correct units in the second box.
e) Find the standard deviation of the sampling distribution of the sample mean.
Put the numeric value rounded to two decimal places in the first box and the correct units in the second
box.
f) What is the shape of the sampling distribution of the sample mean?
Select an answer
Why? Check all that apply:
population is normal
o is unknown
In is less than 30
population is not normal
In is at least 30
o is known
g) Find the probability that the sample mean cholesterol level of the 21 randomly selected women age 45-
59 in Ghana, Nigeria, and Seychelles is less than 4.7 mmol/l.
Round final answer to 4 decimal places.
DO NOT use the rounded standard deviation from part e in this computation.
Use the EXACT value of the standard deviation with the square root.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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