Question

Suppose that quiz scores in a beginning statistics class have a mean of 7.47.4 with a standard deviation of 0.20.2. Using Chebyshev's Theorem, state the range in which at least 88.9%88.9% of the data will reside. Please do not round your answers.

Answer 1

Answer:

(6.8,8)

Step-by-step explanation:

mean=7.4

standard deviation=0.2

we have to find the range in which at least 88.9% of the data will reside

1-1/k²=0.889

1/k²=1-0.889

1/k²=0.111

k²=1/0.111=9.009

k=3.002

so, k=3

The range of values can be computed as mean±k(standard deviation).

Thus, the range in which at least 88.9% of the data will reside is

(mean-k(standard deviation), mean+ k(standard deviation))

(7.4-3(0.2),7.4+3(0.2))

(7.4-0.6,7.4+0.6)

(6.8,8)

Thus, the range in which at least 88.9% of the data will reside is (6.8,8).