The mean output of a certain type of amplifier is 352352 watts with a variance of 121121. If 5757 amplifiers are sampled, what is the probability that the mean of the sample would differ from the population mean by more than 2.32.3 watts

88.56%

Step-by-step explanation:

given:

Mean M = 352

Variance V = 121

sample size n = 57

. To find p(x <2..32 )

for that we use the formula of normal distribution and the normal distribution table

First: The normal distribution formula of Z score is :

z = (x - M ) / (SD / sqrt(n))

. SD = sqrt(variance)

SD = sqrt(121)

SD = 11

On plugging the values we get

z = (352-2.32) / (11/sqrt(57))

z = 1.58

Second, we refer to the Z table for probability value

P(z < 1.58) can be obtained from the Z table

=0.8858.

the answer is a i belaeve atleast thats what my calculator just gfave me

step-by-step explanation:

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.