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Deliverable 02 Questions 918

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Subject
Statistics
School
Rasmussen University
Type
Homework
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Deliverable 02 Worksheet
Instructions: The following worksheet is shown to you by a student who is asking for help. Your
job is to help the student walk through the problems by showing the student how to solve each
problem in detail. You are expected to explain all of the steps in your own words.
Key:
<i> - This problem is an incorrect. Your job is to find the errors, correct the errors, and
explain what they did wrong.
<p> - This problem is partially finished. You must complete the problem by showing all
steps while explaining yourself.
<b> - This problem is blank. You must start from scratch and explain how you will
approach the problem, how you solve it, and explain why you took each step.
1) <p> Assume that a randomly selected subject is given a bone density test. Those tests
follow a standard normal distribution. Find the probability that the bone density score for
this subject is between -1.53 and 1.98
Student’s answer: We first need to find the probability for each of these z-scores using
Excel.
For -1.53 the probability from the left is 0.0630, and for 1.98 the probability from the left is
0.9761.
Continue the solution:
P(-1.53 < z < 1.98)
P(-1.53) = 0.0630
P(1.98) = 0.9761
We need to then subtract the values.
"=norm.s.dist(1.98,true)" - "=norm.s.dist(-1.53,true)"
(0.976148236) - (0.063008364)
=0.9131
=91.31%
There is 91.31%probability that the bone density score for the subject being
between -1.53 and 1.98

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2) <b> The U.S. Airforce requires that pilots have a height between 64 in. and 77 in. If
women’s heights are normally distributed with a mean of 65 in. and a standard deviation
of 3.5 in, find the percentage of women that meet the height requirement.
Answer and Explanation:
3) <i> Women’s pulse rates are normally distributed with a mean of 69.4 beats per minute
and a standard deviation of 11.3 beats per minute. What is the z-score for a woman
having a pulse rate of 66 beats per minute?
Student’s answer:
Let 
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Corrections:
P(64<x<77)
=norm.dist(77,65,3.5,true)=1
=norm.dist(64,65,3.5,true)=0.387548
=1-0.387548
=0.612148135
=0.6121
=61.21%
There are 61.21% of women that meet the height requirement.

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Deliverable 02 – Worksheet Instructions: The following worksheet is shown to you by a student who is asking for help. Your job is to help the student walk through the problems by showing the student how to solve each problem in detail. You are expected to explain all of the steps in your own words. Key: • • • - This problem is an incorrect. Your job is to find the errors, correct the errors, and explain what they did wrong. - This problem is partially finished. You must complete the problem by showing all steps while explaining yourself. - This problem is blank. You must start from scratch and explain how you will approach the problem, how you solve it, and explain why you took each step. 1) Assume that a randomly selected subject is given a bone density test. Those tests follow a standard normal distribution. Find the probability that the bone density score for this subject is between -1.53 and 1.98 Student’s answer: We first need to find the probability for each of these z-scores using Excel. For -1.53 the probability from the left is 0.0630, and for 1.98 the probability from the left is 0.9761. Continue the solution: P(-1.53 < z < 1.98) P(-1.53) = 0.0630 P(1.98) = 0.9761 We need to then subtract the values. "=norm.s.dist(1.98,true)" - "=norm.s.dist(-1.53,true)" (0.976148236) - (0.063008364) =0.9131 =91.31% There is 91.31%probability that the bone density score for the subject being between -1.53 and 1.98 2) The U.S. Airforce requires tha ...
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