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.

answer

explanation

the area of a circle is calculated using the formula

or

we can use any of the above formulae to calculate the required. preferably, let us use the latter since the diameter is

we substitute the diameter into the formula to get:

this is the exact value for the area of the circle.

the correct answer is a

3.76%

step-by-step explanation: