Charles K. answered • 09/08/19
Retired College Professor Working With Students At All Levels
Problem:
Mr. Z gave a test in Physics. Scores for the class are: 25, 53, 70, 33, 26, 71, 54, 31, 51, and 72. Assume the 10 scores from this test are normally distributed.
A. Find the Median, Mean, and Interquartile Range of these 10 values. Round to one decimal place if needed. Indicate which is the mean and which is the median. Show your work!
B. Find the Standard Deviation of these 10 values. Round to one decimal place if needed. Even though this is the entire class, use the “sample” method when finding the Standard Deviation. Show your work! If you use technology to find the Standard Deviation, then explain how to compute it manually, and what technology you used to find it.
C. If this set of scores is representative of all Physics classes Mr. Z has (in other words, with the same mean and same standard deviation), on average what percent of students pass this test? (Passing is 60%, 60/100, or higher.) Use technology or Standard Normal table to find probabilities and show all your reasoning to answer the question. Compute z-score to 3 decimal places and percent probability to 2 decimal places or decimal probability to 4 decimal places
PART A.
Mean.
To find the mean we first sum the numbers and then divide by n = 10
25 + 53 + 70 + 33 + 26 + 71 + 54 + 31 + 51 + 72 = 486
Now we divide by 10 to find the mean.
Mean = 486/10 = 48.6
Mean = 
Median.
To find the median we sort the numbers and arrange them from the smallest to the largest.
25
26
31
33
51
53
54
70
71
72
If there are an odd number of values, we simply take the middle value as the median. However, in this case we have an even number of values. Therefore, we have to use the two middle values. We take the two middle values and take an average of them. This becomes the median. If you look above you will see that I left spaces around the two center values. I’ll use these two numbers and calculate an average.
Median = (51 + 53) / 2
Median = 104 / 2
Median = 52
Median = 
Interquartile Range.
The interquartile range is simply quartile 3 less quartile 1. I repost the data from above and find the median for the first 4 observations and the median for the last 4 observations.
25
26
31
33
51
53
54
70
71
72
Q1 = (26 + 31) / 2
Q1 = 57 / 2
Q1 = 28.5
Q3 = (70 + 71) / 2
Q3 = 141 / 2
Q3 = 70.5
Interquartile Range = IQR = 70.5 – 28.5 = 42
Interquartile Range = 42
PART B.
Standard Deviation.
Sample Variance =
Sample Standard Deviation =
In order to calculate the standard deviation, we first have to calculate the variance. Then the standard deviation is merely the square root of the variance.
The first step is calculating the variance is to subtract the mean from observation.
25 – 48.6 = –23.6
26 – 48.6 = –22.6
31 – 48.6 = –17.6
33 – 48.6 = –15.6
51 – 48.6 = 2.4
53 – 48.6 = 4.4
54 – 48.6 = 5.4
70 – 48.6 = 21.4
71 – 48.6 = 22.4
72 – 48.6 = 23.4
Now, notice that if we sum these values, the sum is equal to zero. Which is why we square these values.
–23.6
–22.6
–17.6
–15.6
2.4
4.4
5.4
21.4
22.4
23.4
0
(–23.6)^2 = 556.96
(–22.6)^2 = 510.76
(–17.6)^2 = 309.76
(–15.6)^2 = 243.36
(2.4)^2 = 5.76
(4.4)^2 = 19.36
(5.4)^2 = 29.16
(21.4)^2 = 457.96
(22.4)^2 = 501.76
(23.4)^2 = 547.56
Sum = 3,182.4
Now, if we go back to the formula, it tells us to divide by (n–1). Since there are 10 observations we divide by 9.
Variance = 3,182.4 / 9 = 353.6
To calculate the standard deviation we take the square root of the variance. In this case, we take the square root of 353.6 which is, 18.8043.
Sample Standard Deviation = 353.6^0.5 = 18.8043 rounded to 4 decimal places
If we use Microsoft Excel and the formulas, =var(range) and =stdev(range) for the variance and standard deviation, we arrive at the same exact values.
Sample Variance = 353.6
Sample Standard Deviation = 18.8043 =
PART C.
This part asks us to figure out what percent of students passed an exam given the mean and standard deviation we just figured out above. The question asks us to use the normal distribution. This is the bell shaped distribution we’re all familiar with.
Mean = 48.6
Standard Deviation = 18.8043
The first thing I do is draw a picture of what I’m looking for. In this case, 60 is the minimum passing grade. So we’re looking for the area under the curve in shaded region.
Z = (x – u) / st.dev
Z = (60 – 48.6) / 18.8043 = 0.61
If you use a Z-table we look up 0.61 on the left and top and find the area under the curve to the right of 60 to be 0.2709. In other words, the area under the curve associated with a passing grade is 27.09 percent. Indeed, if we look at the 10 scores, we see that only three students passed the test with scores of 70, 71 and 72. Which, to be perfectly honest with you, is not surprising for a basic probability and statistics class. I should know. I taught the class for about 20 years every single semester to undergraduates. However, by the end of the semester, most of the students were passing the exams. But it took a lot of office hours and hard work to get the students to learn the material.
Therefore, on average, about 27.09 percent of students pass.
.
