Question

The heights of the adults in one town have a mean of 67.1 inches and a standard deviation of 3.5 inches. What can you conclude from​ Chebyshev's theorem about the percentage of adults in the town whose heights are between 60.1 and 74.1​ inches?

Answer 1

Answer:

The percentage of adults in the town whose heights are between 60.1 and 74.1​ inches

P( 60.1≤X≤74.1) = 2.26 %

Step-by-step explanation:

Step(i):-

Given that the heights of the adults in one town have a mean of 67.1 inches

Mean of the Population = 67.1 inches

The standard deviation of the Population = 3.5inches

Let X₁ =60.1

$Z = \frac{x-mean}{S.D} = \frac{60.1-67.1}{3.5} = -2$

Let X₂ = 74.1

$Z = \frac{x-mean}{S.D} = \frac{60.1-74.1}{3.5} = -4$

Step(ii):-

The probability of adults in the town whose heights are between 60.1 and 74.1​ inches

P( X₁ ≤X≤X₂) = P(Z₁≤Z≤Z₂)

                    = P( -2≤ Z≤-4)

                   = |A(-4) -A(-2)|

                   = A(4) - A(2)|     (∵ A(-Z) =A(Z))

                  = 0.4998 - 0.4772

                 = 0.0226

Final answer:-

The probability of adults in the town whose heights are between 60.1 and 74.1​ inches

P( X₁ ≤X≤X₂) = 0.0226

The percentage of adults in the town whose heights are between 60.1 and 74.1​ inches

P( 60.1≤X≤74.1) = 2.26%