S. (21 points). The weights of the contents of a cereal box are normally weight

Question

S. (21 points). The weights of the contents of a cereal box are normally weight of 20 ounces and a standard deviation of 0 07 ounces a) What is the weight of the contents of a cereal box that is distributed with a 1 standard deviation above the mean 2 standard deviations below the mean b) Compute the z value of the following weights, Round to 2 decimal places Show your work 19.857 19.79 ounces c) The supervisor says that there's at 1% chance or more that the weight of a randomly selected cereal box is less than 19.79 ounces. Is he right or not? Explain your answer. 3pts) d) Boxes in the lower 5% do not meet the minimum weight requirements and must be repackaged. What is the minimum weight requirements for a cereal box?

Explanation / Answer

5. a) Weight of the cereal box that is one standard deviation above mean = 20 + 1x0.07 = 20.07

Weight of cereal box that is two standard deviations above mean = 20 + 2x0.07 = 20.14

b) (x - mean)/standard deviation

When x = 19.857,

Z = (19.857 - 20)/0.07

= -2.04

When x = 20.14,

Z = (20.14 - 20)/0.07

= 2

When x = 19.79,

Z = (19.79 - 20)/0.07

= -3

c) P(X < 19.79) = P(Z < - 3)

= 0.0013

= 0.13%

The supervisor is wrong

d) Let the minimum requirement be M

P(X < M) = 0.05

P(Z < (M - 20)/0.07) = 0.05

(M - 20)/0.07 = 1.645

M = 19.8849 ounces

e) Let the range be (E,F)

P(X < E) = (1-0.92)/2

= 0.04

P(Z < (E - 20)/0.07) = 0.04

(E - 20)/0.07 = -1.75

E = 19.8775

(F - 20)/0.07 = 1.75

F = 20.1225

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