Lloyd\'s Cereal company package cereal in I pound boxes (16 ounces). A sample of

Question

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? 0.0124 0.3773 02062 0.9938 00062 None of the above in a large population. 82% of the households have cable to, A simple random sample of 225 households it to he contacted and the sample proportion computed. What is the mean and standard deviation of the sampling distribution of the sample proportions? mean - 184.50, standard deviation = 0025 mean = 0.82. standard deviation = 00007 mean = 18430, standard deviation =00007 mean - 0.82, standard deviation = 0.025 mean = 0.82, standard deviation = 0.5656

Explanation / Answer

1)

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)

2)

P(having cable tv) = 0.82

n =225

mean of sample proportion = np = (225/225)*0.82 = 0.82

varinace = np(1-p) = (225)*0.82*0.18 = 33.21

standard deviation = 5.763

standard deviation of sample proportion = 5.763/225 = 0.0256

so answer is (d)

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