Question

The total cholesterol values for a certain population are approximately normally distributed with a mean of 200 ??/100?? and a standard deviation of 20 ??/100??. Find the total cholesterol values corresponding to the first quartile, the second quartile and the third quartile for this population.

Answer #1

The total cholesterol values for a certain population are
approximately normally distributed with a mean of mg/100ml and a
standard deviation of 20mg/100ml.
Find thetotalcholesterol valuescorresponding to the first
quartile,thesecond quartileand the third quartilefor
thispopulation.

The weights of broilers (commercially raised chickens ) are
approximately normally distributed with mean 1387 grams and
standard deviation 161 grams . What is the probability that a
randomly selected broiler weighs more than 1,532 grams ? Write only
a number as your answer Round to 4 decimal places ( for example
0.0048 ). Do not write as a percentage .
The weights of broilers (commercially raised chickens) are
approximately normally distributed with mean 1387 grams and
standard deviation 161...

a.) A sample set is normally distributed with a mean of 2.8 and
a standard deviation of 0.7. Approximately, what percentage of the
sample is between the values of 2.1 and 3.5?
b.) A sample set is normally distributed with a mean of 66 and a
standard deviation of 8.
Find a z-score corresponding to a given value of 82.

The amount of rainfall in January in a certain city is normally
distributed with a mean of
3.9
inches and a standard deviation of
0.4
inches. Find the value of the quartile
Q1
Round to the nearest tenth.
A.1.0
B.4.2
C.3.6
D.3.7

Total cholesterol levels for women under 30 in Podunk are
normally distributed with mean 173 mg/dL and standard deviation
27.2 mg/dL.
a. If a woman is selected at random, find the probability that
her total cholesterol is higher than 180 mg/dL.
b. What cholesterol levels make up the top 70% of this
distribution?

A.) Suppose a population is known to be normally distributed
with a mean, μ, equal to 116 and a standard deviation, σ, equal to
14. Approximately what percent of the population would be between
116 and 144?
B.) Suppose a population is known to be normally distributed
with a mean, μ, equal to 116 and a standard deviation, σ, equal to
14. Approximately what percent of the population would be between
116 and 130?
C.) Suppose a population is known...

20. Lifetimes of a certain brand of tires are approximately
normally distributed with mean 40,000 miles and standard deviation
2,500 miles. If the company making the tires did not want to
replace more than 3% of the tires, what is the lowest mileage the
company should make for the warranty? Show all work do not
round.

1. Suppose a population is known to be normally distributed with
a mean, μ, equal to 116 and a standard deviation, σ, equal to 14.
Approximately what percent of the population would be between 102
and 144?
2. Suppose a population is known to be normally distributed with
a mean, μ, equal to 116 and a standard deviation, σ, equal to 14.
Approximately what percent of the population would be between 102
and 130?
3. Suppose a population is known...

The cholesterol content of large chicken eggs is normally
distributed with a mean of 187 milligrams and standard deviation 15
milligrams. Use this information to answer the questions below.
Round your answer to two decimal places.
a. What is the probability one of these eggs is between 186 and
200?
b. Suppose six eggs are randomly chosen from this population of
eggs. What is the probability the average milligrams of cholesterol
of the sample be less than 200 milligrams?
c....

The weights of a certain dog breed are approximately normally
distributed with a mean of 53 pounds, and a standard deviation of
5.9 pounds. Answer the following questions. Write your answers in
percent form. Round your answers to the nearest tenth of a
percent.
a) Find the percentage of dogs of this breed that weigh less than
53 pounds. %
b) Find the percentage of dogs of this breed that weigh less than
49 pounds. %
c) Find the percentage...

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