nat is the probability that sample will be larger than 670 mg? 2. In a study of the life expectancy of 400 people in a certain geographic region, the mean age at death was 70 years, 'and the standard deviation was 5.1 years. If a sample of 50 people from this region is selected, what is the probability that the mean life expectancy will be less than 68 years?
nat is the probability that sample will be larger than 670 mg? 2. In a study of the life expectancy of 400 people in a certain geographic region, the mean age at death was 70 years, 'and the standard deviation was 5.1 years. If a sample of 50 people from this region is selected, what is the probability that the mean life expectancy will be less than 68 years?
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Answer no.2 with on point answer and clear solutions the example is on the second pic
Transcribed Image Text:is the probability that
ble will be larger than 670 mg?
2. In a study of the life expectancy of 400 people in a certain geographic region, the
mean age at death was 70 years, and the standard deviation was 5.1 years. If a
sample of 50 people from this region is selected, what is the probability that the
mean life expectancy will be less than 68 years?
3. The average cholesterol content of a certain canned goods is 215 milligrams, and
the standard deviation is 15 milligrams. Assume that the variable is normollh
distributod If
Transcribed Image Text:Solution for #1.a:
Step1: Identify the parts of the problem.
Given: u= 46.2 minutes; o= 8 minutes; X= 43 minutes;
Find: P(X < 43)
n = 50 students
%3D
Step 2: Use the formula to find the z-score.
Solution:
43 - 46.2
8
Given: u= 1200 hours;
o = 250 hours;
Vn
V50
X = 1 150 & 1 250 hours
n = 100 bulbs
Unknown: P(1150 < X < 1250)
z = -2.83
Step 2: Use the formula to find the z-score.
Step 3: Use the z-table to look up the z-score you calculated in step 2.
z = -2.83 has a corresponding area of 0.4977
1150 1200
1250 1200
250
Vn
V100
250
Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade
V100
the part that you're looking for: P(X < 43)
z = -2
Z = 2
Step 3: Use the z-table to look up the z-score you calculated in step 2.
z = +2 has a corresponding area of 0.4772
Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade
the part that you're looking for: P(1150 <X < 1250)
sha
ed part
0.4977
-3
-2
-1
1
2
3
-2.83
0.4772
0.4772
Since we are looking for the probability less than 43 minutes, the shaded part
will be on the left part of - 2.83.
-3
-2
-1
2
3
Since we are looking for the probability between 1 150 hours and 1 250 hours,
the shaded part will be between -2 and 2.
Step 5: Subtract your z-score from 0.500.
P(X < 43) = 0.500 0.4977
P(X < 43) = 0.0023
Step 5: Add the two z-score values.
P(1150 <X < 1250) = 0.4772 + 0.4772
P(1150 <X < 1250) = 0.9544
%3D
Step 6: Convert the decimal in Step 5 to a percentage.
P(X < 43) = 0.23%
%3D
Step 6: Convert the decimal in Step 5 to a percentage.
P(1150 <X < 1250) = 95.44%
%3D
. Therefore, the probability that a randomly selected 50 senior high school
students will complete the examination in less than 43 minutes is 0.23%. No, it's not
reasonable since the probability is less than 1.
. Therefore, the probability of randomly selected 100 bulbs to have a sample
mean between 1 150 hours and 1 250 hours is 95.44%.
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