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Ust Need Correction On Statistics Paper-Tutoring On The Normal Distribution
Inferential Statistics and Analytics
Deliverable 02 –Tutoring on the Normal Distribution
Rasmussen College
11/13/17
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 -1.93 and 2.37
Student’s answer: We first need to find the probability for each of these z-scores using Excel.
For -1.93 the probability from the left is 0.0268, and for 2.37 the probability from the left is 0.9911.
Continue the solution:
We first need to find the probability for each of these z-scores using Excel.
For -1.93 the probability from the left is 0.0268, and for 2.37 the probability from the left is 0.9911.
Therefore, the probability that the bone density score that is
P (-1.93 < Z < 2.37) = 0.0268 – 0.9911.
= -0.9643.
= 0.9643.
2) 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 63.8 in. and a standard deviation of 2.6 in, find the percentage of women that meet the height requirement.
Answer and Explanation:
P (64 Normal distribution µ = 63.8, α = 2.6 We convert this to standard normal using the formula. z=(x- µ)/α Z1 = (64- 63.8)/2.6 = 0.076923 = 0.08 Z2 = (77 – 63.8)/2.6 = 5.076923 = 5.08 Therefore P (0.08 < Z< 5.08 area in between 0.08 and 5.08 P (64 = P (Z<5.08) – P (Z<0.08) = 1.0000 – 0.5319 = 0.4681. 3) 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: Mean = μ = 69.4 bpm Standard deviation = σ = 11.3 bpm To solve any problem involving a normal distribution we need to calculated Z and we are requiring the Z tables. X - μ Z = ----------- σ 66– 69.4 Z = ----------- 11.3 = -0.30088 From the tables table of Z values we get = 1.54. 4) What is the cumulative area from the left under the curve for a z-score of -1.645? What is the area on the right of that z-score? Answer and Explanation: We use the Z- score tables Cumulative area from the left under the curve for a z score of -1.645 = separate the score that is -1.6 and 0. 045 = 0. 0505 The area to the right = 1- area to the left. = 1- 0.0505 = 1.9495 5) If the area under the standard normal distribution curve is 0.8980 from the right, what is the corresponding z-score? Student’s answer: We plug in “=NORM.INV(0.8980, 0, 1)” into Excel and get an area of 1.27. Corrections: To find the corresponding z-score we need to use the z-score tables Look inside the table the value of 0.8980 = 1.27 6) 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 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 inches. Continue the solution: We need to look for the probability that the head breadth of a man is less than 6.2 in. We use the formula z= x-μ _______ σ To convert from the non-standard normal distribution to the standard normal distribution. 6.2 in is converted as follows Z = x-μ _______ σ = _______________________________6.2-6.0__________________________________ 1 .0 = 0.2 We find the cumulative area to the left of z= 0.2 which is an area of 0.5793 Thence the percentage of the men will fill into a manhole with this diameter = 57.93%
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