Answer:
The probability of the widget weights lie between 43 and 87 ounces
P(43 ≤ x ≤ 87 ) = 0.7605
The percentage of the widget weights lie between 43 and 87 ounces
P(43 ≤ x ≤ 87 ) = 76 %
Explanation:
Step(i):-
Given mean of the Population = 50ounces
Given standard deviation of the Population = 10 ounces
Let 'X' be the random variable in Normal distribution
Given X = 43
Given X = 87
Step(ii):-
The probability of the widget weights lie between 43 and 87 ounces
= |A(3.7) + A(-0.7)|
= |A(3.7) + A(0.7)| (∵A(-z) = A(z))
= 0.4994 + 0.2611
= 0.7605
Final answer:-
The probability of the widget weights lie between 43 and 87 ounces
P(43 ≤ x ≤ 87 ) = 0.7605
The percentage of the widget weights lie between 43 and 87 ounces
P(43 ≤ x ≤ 87 ) = 76 %