Lloyd's cereal company package cereal in i pound boxes (16 ounces). a sample of 64 boxes is selected at random from the production line even hour, and if the average weight is less than 15 ounces, the machine is adjusted to increase the amount of cereal dispensed. if the mean for i hour is 1 pound and the standard deviation is 02 pound, what is the probability that the amount dispensed per box will have to be increased?
a. 0.0124
b. 0.3773
c. 02062
d. 0.9938
e. 00062
f. none of the above

e

Step-by-step explanation:

mean = 16 ounces

standard deviation = 0.2×16 ounces = 3.2 ounces

standard deviation of sample mean = 3.2/√64 = 3.2/8 =0.4

z value of 15 is (15-16)/0.4 = -1/0.4 = -2.5, using z-table corresponding p value = 0.00621

P(average <15) =0.00621

so answer is (e)

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$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p