Question

Find the indicated probability: Assume that the weights of candies are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept candies that are weighing between 5.48 g and 5.82 g. What percentage of candies will be rejected by the machine? Give your answer in the percentage format (using % symbol), rounded to two decimal places. HINT: Percentage = probability = area under the curve; Percentage rejected = 100% – percentage accepted

98.04% is not the correct answer.

Answer 1

Solution :

Given that ,

mean = = 5.67

standard deviation = = 0.070

P(5.48 < x < 5.82) = P[(5.48 - 5.67)/ 0.070 ) < (x - ) /  < (5.82 - 5.67) / 0.070 ) ]

= P(-2.71 < z < 2.14)

= P(z < 2.14) - P(z < -2.71)

Using z table,

= 0.9838 - 0.0034

= 0.9804

= 1 - 0.9804

= 0.0196

Percentage rejected = 1.96%