The mean output of a certain type of amplifier is 466 watts with a standard deviation of 10 watts. If 32 amplifiers are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 1.6 watts? Round your answer to four decimal places.

0.8186

Step-by-step explanation:

The formula to use to solve this question is given as:

z = (x-μ)/σ/√n, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

n is the random number of samples

P(z < (x-μ)/σ/√n)

(x-μ) = 1.6 watts

σ = 10 watts

n = 32 amplifiers

Hence,

P(z < 1.6/10/√32)

P(z < 1.6/1.767766953)

P(z < 0.9050966799)

Approximately to 2 decimal places)

P(z < 0.91)

= 0.81859

Rounded to four decimal places, the probability that the mean of the sample would differ from the population mean by less than 1.6 watts is 0.8186.

c

step-by-step explanation:

i think because what else can be a sample of every member

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.