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

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step-by-step explanation:

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