Question

Suppose that quiz scores in a beginning statistics class have a mean of 6.4 with a standard deviation of 0.6. Using Chebyshev's Theorem, state the range in which at
least 75% of the data will reside. Please do not round your answers.
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Answer 1

(5.2,7.6)

mean=6.4

standard deviation=0.6

we have to find the range in which at least 75% of the data will reside

1-1/k²=0.75

1/k²=1-0.75

1/k²=0.25

k²=1/0.25=4

k=2

so, k=2

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

Thus, the range in which at least 75% of the data will reside is

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

(6.4-2(0.6),6.4+2(0.6)

(6.4-1.2,6.4+1.2)

(5.2,7.6)

learn more of standard deviation here brainly.com/question/14671301

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