Demonstrate the use of the normal distribution, the standard normal distribution, and the central limit theorem for calculating areas under the normal curve and exploring these concepts in real life applications.
Frank has only had a brief introduction to statistics when he was in high school 12 years ago, and that did not cover inferential statistics. He is not confident in his ability to answer some of the problems posed in the course.
As Frank's tutor, you need to provide Frank with guidance and instruction on a worksheet he has partially filled out. Your job is to help him understand and comprehend the material. You should not simply be providing him with an answer as this will not help when it comes time to take the test. Instead, you will be providing a step-by-step breakdown of the problems including an explanation on why you did each step and using proper terminology.
What to Submit
To complete this assignment, you must first download the word document, and then complete it by including the following items on the worksheet:
Correct any wrong answers. You must also explain the error performed in the problem in your own words.
Partially Finished Work
Complete any partially completed work. Make sure to provide step-by-step instructions including explanations.
Show how to complete any blank questions by providing step-by-step instructions including explanations.
Your step-by-step breakdown of the problems, including explanations, should be present within the word document provided. You must also include an Excel workbook which shows all your calculations performed.
Explanation & Answer
Hello, kindly find the attached documents. In case of any question feel free to ask. Thank you
Running Head: NORMAL DISTRIBUTION
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
For -1.53 the probability from the left is 0.0630, and for 1.98 the probability from the left is
P (z ≤ −1.53) = 0.0630
The probability is the shaded area on the graph below.
P(z ≤ 1.98) = 0.9761
The probability is represented by the shaded area on the graph below.
To obtain the probability that the density score is between -1.53 and 1.98 we find
the difference between the two areas.
P(−1.53 ≤ z ≤ 1.98) = 0.9761 − 0.0630 = 0.9131
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏(−𝟏. 𝟓𝟑 ≤ 𝐳 ≤ 𝟏. 𝟗𝟖) = ...