Question

The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 7.8 ounces and standard deviation 0.2 ounces. (a) What is the probability that the average weight of a bar in a random sample with three of these chocolate bars is between 7.64 and 7.96 ounces?

ANSWER:

(b) For a random sample of three of these chocolate bars, what is the level L such that there is a 4% chance that the average weight is less than L?

ANSWER:

Shelia's doctor is concerned that she may suffer from gestational diabetes (high blood glucose levels during pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. A patient is classified as having gestational diabetes if the glucose level is above 140 milligrams per deciliter one hour after a sugary drink is ingested. Shelia's measured glucose level one hour after ingesting the sugary drink varies according to the Normal distribution with mean 129 mg/dl and standard deviation 8 mg/dl. Let LL denote a patient's glucose level.

(a) If measurements are made on three different days, find the level LL such that there is probability only 0.05 that the mean glucose level of three test results falls above LL for Shelia's glucose level distribution. What is the value of LL?
ANSWER:

(b) If the mean result from the three tests is compared to the criterion 140 mg/dl, what is the probability that Shelia is diagnosed as having gestational diabetes?
ANSWER:

The scores of students on the SAT college entrance examinations at a certain high school had a normal distribution with mean μ=559.2μ=559.2 and standard deviation σ=28σ=28.

(a) What is the probability that a single student randomly chosen from all those taking the test scores 563 or higher?
ANSWER:  

For parts (b) through (d), consider a random sample of 25 students who took the test.

(b) What are the mean and standard deviation of the sample mean score x¯x¯, of 25 students?
The mean of the sampling distribution for x¯x¯ is:  
The standard deviation of the sampling distribution for x¯x¯ is:

(c) What z-score corresponds to the mean score x¯x¯ of 563?
ANSWER:

(d) What is the probability that the mean score x¯x¯ of these students is 563 or higher?
ANSWER:

Answer 1

1)a) P(7.64 < < 7.96)

= P((7.64 - < ( -   < (7.96 - ))

= P((7.64 - 7.8)/(0.2/) < Z < (7.96 - 7.8)/(0.2/))

= P(-1.39 < Z < 1.39)

= P(Z < 1.39) - P(Z < -1.39)

= 0.9177 - 0.0823

= 0.8354

B) P( < L) = 0.04

Or, P( < ) = 0.04

Or, P(Z < (L - 7.8)/(0.2/)) = 0.04

Or, (L - 7.8)/(0.2/) = -1.75

Or, L = -1.75 * (0.2/) + 7.8

Or, L = 7.6

2)a) P( < L) = 0.05

Or, P(( < ) = 0.05

Or, P(Z < (L - 129)/(8/)) = 0.05

Or, (L - 129)/(8/) = -1.645

Or, L = -1.645 * (8/) + 129

Or, L = 121.4

B) P( > 140)

= P( > )

= P(Z > (140 - 129)/(8/))

= P(Z > 2.38)

= 1 - P(Z < 2.38)

= 1 - 0.9913

= 0.0087

3) P(X > 563)

= P((X - )/> (563 - )/)

= P(Z > (563 - 559.2)/28)

= P(Z > 0.14)

= 1 - P(Z < 0.14)

= 1 - 0.5557

= 0.4443

B) = 559.2

= = 28/ = 5.6

C) Z-score =

= (563 - 559.2)/5.6 = 0.68

D) P(> 563)

P(Z > 0.68)

= 1 - P(Z < 0.68)

= 1 - 0.7517

= 0.2483