The lengths of a lawn mower part are approximately normally distributed with a given mean Mu = 4 in. and standard deviation Sigma = 0.2 in. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.? 34% 68% 95% 99.7%

b) 68%

The percentage of the parts will have lengths between 3.8 in. and 4.2 in

P( 3.8 ≤ x ≤ 4.2) = 0.6826 = 68 %

Step-by-step explanation:

Let 'X' be the normally distributed

mean 'μ'= 4 inches

standard deviation 'σ' = 0.2 inches

Case(i):-

when x₁ = 3.8 inches

Case(ii):-

when  x₂= 3.8 inches

The probability of the parts will have lengths between 3.8 in and 4.2 in

                        = P(Z≤1) - P(Z≤-1)

                        =  0.5 +A(1) -(0.5-A(1)

                       = 2 A(1)

                       = 2×0.3413

                      = 0.6826

Conclusion:-

The percentage of the parts will have lengths between 3.8 in. and 4.2 in

P( 3.8 ≤ x ≤ 4.2) = 0.6826 = 68 %

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

what i would say the best answer is a. 40.5