Answer:
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
[tex]Z_{1} = \frac{x_{1}-mean }{S.D} = \frac{3.8-4}{0.2} = -1[/tex]
Case(ii):-
when x₂= 3.8 inches
[tex]Z_{2} = \frac{x_{2}-mean }{S.D} = \frac{4.2-4}{0.2} = 1[/tex]
The probability of the parts will have lengths between 3.8 in and 4.2 in
[tex]P( 3.8\leq x\leq 4.2) = P(-1\leq z\leq 1)[/tex]
= 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 %
