Mrs. Jones algebra 2 class scored very well on yesterday's quiz. With one exception, everyone received an A. Within how many standard deviations from the mean do all the quiz grades fall ? Ó =3.1
95, 92,94,88,96,99,91,93,94,97,95,97
font problems. Is that supposed to be sigma = 3.1?
If so, just figure the mean, and find the maximum difference of each grade from the mean. Then divide that by 3.1 to get the number of std.
To find within how many standard deviations from the mean do all the quiz grades fall, we can follow these steps:
Step 1: Find the mean (average) of the quiz grades.
Step 2: Calculate the standard deviation of the quiz grades.
Step 3: Determine the range within which all grades fall.
Let's calculate it step-by-step:
Step 1: Find the mean (average).
To find the mean, sum up all the quiz grades and divide by the total number of grades:
95 + 92 + 94 + 88 + 96 + 99 + 91 + 93 + 94 + 97 + 95 + 97 = 1141
Mean = 1141 / 12 = 95.08 (rounded to two decimal places).
Step 2: Calculate the standard deviation.
To calculate the standard deviation, we need to find the squared difference between each grade and the mean. Then, average these squared differences and take the square root. Here's the calculation:
(95 - 95.08)^2 = 0.0064
(92 - 95.08)^2 = 9.6864
(94 - 95.08)^2 = 1.1664
(88 - 95.08)^2 = 50.2464
(96 - 95.08)^2 = 0.8464
(99 - 95.08)^2 = 15.3664
(91 - 95.08)^2 = 16.4864
(93 - 95.08)^2 = 4.3264
(94 - 95.08)^2 = 1.1664
(97 - 95.08)^2 = 3.6764
(95 - 95.08)^2 = 0.0064
(97 - 95.08)^2 = 3.6764
Sum of squared differences = 106.7968
Average of squared differences = 106.7968 / 12 = 8.89973 (rounded to five decimal places)
Now, take the square root of the average:
√8.89973 ≈ 2.9827 (rounded to four decimal places)
Step 3: Determine the range within which all grades fall.
In this case, we need to find out how many standard deviations each grade is from the mean. The formula is:
Standard deviations = (Grade - Mean) / Standard Deviation
Let's calculate it for each grade:
95: (95 - 95.08) / 2.9827 ≈ -0.0268
92: (92 - 95.08) / 2.9827 ≈ -1.0303
94: (94 - 95.08) / 2.9827 ≈ -0.3636
88: (88 - 95.08) / 2.9827 ≈ -2.3819
96: (96 - 95.08) / 2.9827 ≈ 0.3090
99: (99 - 95.08) / 2.9827 ≈ 1.3125
91: (91 - 95.08) / 2.9827 ≈ -1.3718
93: (93 - 95.08) / 2.9827 ≈ -0.6969
94: (94 - 95.08) / 2.9827 ≈ -0.3636
97: (97 - 95.08) / 2.9827 ≈ 0.9796
95: (95 - 95.08) / 2.9827 ≈ -0.0268
97: (97 - 95.08) / 2.9827 ≈ 0.9796
All grades fall within approximately -2.3819 to 1.3125 standard deviations from the mean.
To determine within how many standard deviations from the mean all the quiz grades fall, you need to calculate the mean of the quiz grades and then find the range in which all the grades exist based on the standard deviation.
1. Find the mean (average) of the quiz grades:
- Add up all the quiz grades: 95 + 92 + 94 + 88 + 96 + 99 + 91 + 93 + 94 + 97 + 95 + 97 = 1141
- Divide the sum by the number of grades: 1141 / 12 = 95.08 (rounded to 2 decimal places)
- The mean of the quiz grades is approximately 95.08.
2. Calculate the range within which all the quiz grades exist based on the standard deviation:
- The standard deviation (σ) given is 3.1.
- Subtract the mean from the smallest grade: 88 - 95.08 ≈ -7.08
- Divide the result by the standard deviation: -7.08 / 3.1 ≈ -2.28
- Subtract the mean from the largest grade: 99 - 95.08 ≈ 3.92
- Divide the result by the standard deviation: 3.92 / 3.1 ≈ 1.27
- The range within which all the quiz grades fall is approximately -2.28 to 1.27 standard deviations from the mean.
Therefore, all the quiz grades in Mrs. Jones' algebra 2 class fall within approximately -2.28 to 1.27 standard deviations from the mean.