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.
Delegate your assignment to our experts and they will do the rest.
<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.
<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.
P ( = 0.0630 P ( = 0.9761
Continue the solution:
Finish the problem giving step-by-step instructions and explanations.
Probability that the bone density score is less than -1.53 is 0.0630
P ( = NORMSDIST (-1.53) = 0.0630
The probability that bone density score is less than 1.98 is 0.9761
P ( = NORMSDIST (1.98) = 0.9761
The probability that the bone density falls between -1.53 and 1.98 is the difference between the two probabilities
P (-1.53 = P ( - P (
= 0.9761 – 0.0630 = 0.9131
The probability that the bone density score for this subject is between -1.53 and 1.98 is 0.9131
<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:
Enter your step-by-step answer and explanations here.
We find the probability the height is less than 64
P (x ) = NORM.DIST (64, 65, 3.5, TRUE) = 0.3875
We find the probability the height is less than 77
P (x ) = NORM.DIST (77, 65, 3.5, TRUE) = 0.9997
The probability that height is between 64 and 77 is the difference between the two probabilities
P (64 = P (x ) - P (x ) = 0.9997- 0.3875 = 0.6122
Convert the probability into percentage.
61.22% of women meet the height requirement.
<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
Corrections:
Let
Where x is beats per minute, and is the standard deviation
x = 66, = 69.4 and 11.3
<b> What is the cumulative area from the left under the curve for a z-score of -0.875? What is the area on the right of that z-score?
Answer and Explanation:
Enter your step-by-step answer and explanations here.
Cumulative area to the left is P (z -0.875)
P (z -0.875) = NORMSDIST (-0.875) = 0.1908
Area on the right of z-score is P (z > -0.875)
P (z > -0.875) = 1 - P (z -0.875)
P (z > -0.875) = 1- 0.1908 = 0.8092
<i> If the area under the standard normal distribution curve is 0.6573 from the right, what is the corresponding z-score?
Student’s answer : We plug in “=NORM.INV (0.6573, 0, 1)” into Excel and get a z-score of 0.41.
Corrections:
Enter your corrections and explanations here.
Let the associated z-score be c
Probability normal distribution to the right
P (z > c) = 0.6573
We find the probability of normal distribution to the left P (z )
P (z ) = 1 - P (z > c) = 1 - 0.6573 = 0.3427
P (z ) = 0.3427
To find c we plug in NORMSINV (0.3427)
The corresponding z-score = -0.41
)
<p> Manhole covers must be a minimum of 22 in. in diameter, but can be as much as 60 in. Men have shoulder widths that are normally distributed with a mean of 18.2 and a standard deviation of 2.09 in. Assume that a manhole cover is constructed with a diameter of 22.5 in. What percentage of men will fit into a manhole with this diameter?
Student’s answer: We need to find the probability that men will fit into the manhole. The first step is to find the probability that the men’s shoulder is less than 22.5 inches.
Enter your step-by-step answer and explanations here.
Only men with less than 22.5 inches can fit into the manhole
P (x < 22.5)
X=22.5, ,
P (x < 225) = NORM.DIST (22.5, 18.2, 2.09, TRUE)
P (x < 22.5) = 0.9802
Convert the probability into percentage
We conclude that 98.02% of men have shoulders width of less than 22.5 inches and will fit into a manhole
Continue the solution:
