Question

3. The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly.

a. Find the probability that the sum of the 40 values is greater than 7,550. (Round your answer to four decimal places.

b. Find the sum that is 1.6 standard deviations below the mean of the sums. (Round your answer to two decimal places.)

4. A researcher measures the amount of sugar in several cans of the same soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100.
a. Find the probability that the sum of the 100 values is greater than 3,909. (Round your answer to four decimal places.)

b. Find the probability that the sum of the 100 values falls between the numbers 3900 and 3910. (Round your answer to four decimal places.

5. An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest.
a. What is the mean of ΣX? (Enter an exact number as an integer, fraction, or decimal.)

b. What is P(Σx < 280)? (Round your answer to five decimal places.)

10. A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.

a. Find the probability that the mean actual weight for the 100 weights is greater than 24.8. (Round your answer to four decimal places.

b. Find the 95^th percentile for the mean weight for the 100 weights. (Round your answer to two decimal places.)

c. Find the 85^th percentile for the total weight for the 100 weights. (Round your answer to two decimal places.

Answer 1

3) a)

Σ X Normal ( nµ , √ n σ_x )

X ~ N ( µ = 7200 , σ = 126.4911 )
P ( X > 7550 ) = 1 - P ( X < 7550 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 7550 - 7200 ) / 126.4911
Z = 2.766
P ( ( X - µ ) / σ ) > ( 7550 - 7200 ) / 126.4911 )
P ( Z > 2.766 )
P ( X > 7550 ) = 1 - P ( Z < 2.766 )
P ( X > 7550 ) = 1 - 0.99720
P ( X > 7550 ) = 0.0028

b)

Given that ,

= 180

= 20

A sample of size n = 40 is taken from this population.

The distribution of the sample sum has the mean and standard deviation ad

= n* = 40 * 180 = 7200

=   = 20*40 = 126.49

The value of sum which is one standard deviation below the mean is   - (1*)

- (1.6*) = 7200 - (1.6* 126.49) = 6997.61