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.

the answer is; biodiversity resulting from few ancestors

adaptive radiation occurs when a population is subjected to a large natural environment with spatial ecological variations. the parts of the population in the different ecologies begin to adapt to their local environments. due to non-random mating across the large populations, this causes the different populations from the large population to speciate within their local environments. they, therefore, diverge to different species sharing a common ancestor.