Question

Bags of whole coffee beans are filled automatically on a production line. A machine fills each bag so that the weight of coffee beans inside is normally distributed with a mean of 290 grams. The label on the bag, however, states that the weight of coffee beans inside is 283 grams.

a. What is the standard deviation of bags of coffee beans, if 13% of the bags have a weight below what is stated on the label?

b. New management wants to be more accurate to their customers and reduce the number of bags that are sent out under the weight of 283 grams listed on the label. They set a goal of sending no more than 1% of bags of coffee that are under the weight of 283 grams. To do this, the management ordered a new filling machine which decreased the standard deviation to 2.3151 grams. The weight of the bags of coffee beans will still be normally distributed. To what mean weight should the new equipment be set, with this new standard deviation and to meet their goal? The machine may only take a one decimal approximation.

Answer 1

Solution:-

Given that,

mean = = 290 grams

x = 283 grams

a) Using standard normal table,

P(Z < z) = 13%

= P(Z < z) = 0.13  

= P(Z < -1.13 ) = 0.13

z = -1.13

Using z-score formula,

x = z * +

283 = -1.13 * + 290

= 283 - 290 / -1.13

= 6.1947

b) = 2.3151

Using standard normal table,

P(Z < z) = 1%

= P(Z < z) = 0.01  

= P(Z < -2.33 ) = 0.01

z = -2.33

Using z-score formula,

x = z * +

283 = -2.33 * 2.3151 +

= 283 - ( -2.33 * 2.3151)

= 288.4 grams.