Question

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%

Answer 1

B

Explanation:

E2020

Answer 2

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 %

Explanation:

Let 'X' be the normally distributed

mean 'μ'= 4 inches

standard deviation 'σ' = 0.2 inches

Case(i):-

when x₁ = 3.8 inches

`Z_(1) = (x_(1)-mean )/(S.D) = (3.8-4)/(0.2) = -1`

Case(ii):-

when x₂= 3.8 inches

`Z_(2) = (x_(2)-mean )/(S.D) = (4.2-4)/(0.2) = 1`

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

`P( 3.8\leq x\leq 4.2) = P(-1\leq z\leq 1)`

= 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 %