The mean of the quiz grades is (91+92+94+88+96+99+91+93+94+97+95+97) / 12 = 94.33.
To find the standard deviation, we first find the variance.
Variance = Σ(xi - mean)^2 / n
Variance = [(91 - 94.33)^2 + (92 - 94.33)^2 + (94 - 94.33)^2 + (88 - 94.33)^2 + (96 - 94.33)^2 + (99 - 94.33)^2 + (91 - 94.33)^2 + (93 - 94.33)^2 + (94 - 94.33)^2 + (97 - 94.33)^2 + (95 - 94.33)^2 + (97 - 94.33)^2] / 12
Variance = (10.7889 + 7.4489 + 0.1089 + 36.5089 + 2.0889 + 20.4889 + 10.7889+ 3.7689 + 0.1089 + 6.8689 + 0.4225 + 6.1069) / 12
Variance = 5.4557
Standard Deviation = √Variance = √5.4557 = 2.3335
All the quiz grades fall within 2 standard deviations of the mean.
Mrs. Jones Algebra 2 class scored very well on yesterday’s quiz. With one exception, everyone received an A. Within how many standard deviations of the mean do all the quiz grades fall? 91, 92, 94, 88, 96, 99, 91, 93, 94, 97, 95, 97 (1 point) Responses 2 2 1 1 3 3 4
1 answer