1 )The first exam grades of last semester’s Stats class had a mean of 67 with a
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Question
1 )The first exam grades of last semester’s Stats class had a mean of 67 with a standard deviation of 5 points. Answer the following questions. Answer the following questions: Jenny received a score of 80. What is her z-score? Is it positive or negative? Why? Michael’s z-score was a -1.5. Did he do better or worse than the average? How can you tell? And by how many points? Anne’s z-score was a 0. What does this tell you about her score in relation to the mean? John received a score of 65. What is his z-score? Is it positive or negative? Why? 2) For the students in the first part of this question (Jenny, Michael, Anne, and John) . 3) The deviation-method formula calculating the standard deviation (and variance) includes the mean. Explain why it makes sense for this measure of the variability of a sample to be calculated using the mean. (Hint: what is it that the standard deviation measures?
Explanation / Answer
Mean = 67
Standard deviation = 5
Z = (X - mean)/standard deviation
Jenny's score = 80
Jenny's z score = (80-67)/5 = 2.6
The score is positive because it is above the mean
Michael's z-score is -0.15
Negative score indicates that his score is less than that of mean. So, he did worse than the average
Michaels score = -1.5x5 + 67 = 59.5
Anne's z score = 0 mean her score is equal to the mean, that is 67
John's z score = (65-67)/5 = -0.4
The z score is negative because John's score is less than the mean
The standard dev iation is calculated using the deviation from mean. The standard deviation is the average of squared deviation from the mean. So, the value that we get indicates how far from the mean, the data shows variation. The mean represents the average and the measure of variability is to find the average variation from the central tendancy. So, it makes sense to use the mean here as the measure of central tendancy.
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