The mean output of a certain type of amplifier is 102 watts with a standard deviation of 12 watts. If 63 amplifiers are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 3.4 watts

0.97547

Step-by-step explanation:

Given That:

Mean (m) = 102

Standard deviation (σ) = 12

Sample size (n) = 63

Probability that sample mean would differ from Population mean by less than 3.4 watts

P(m - s < X < m + s)

P(102 - 3.4 < Z < 102 + 3.4)

Using the Z formula :

Zscore = (x - m) / (σ/√n)

x = 102 - 3.4 = 98.6 ; x = 102 + 3.4 = 105.4

Z = (98.6 - 102) / (12/√63) = −2.248888

P(Z < - 2.2488) = 0.012263 (Z probability calculator)

Z = (105.4 - 102) / (12/√63) = 2.2488886

P(Z < 2.2488) = 0.98774 (Z probability calculator)

0.98774 - 0.012263 = 0.97547

astronomers is your final answer! : )

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)