The mean output of a certain type of amplifier is 498 watts with a standard deviation of 12 watts. If 86 amplifiers are sampled, what is the probability that the mean of the sample would be less than 494.7 watts

Respuesta :

Mean (m) = 498

Standard deviation (σ) = 12

Sample size (n) = 86

The probability that the sample mean would differ from Population mean by less than 494.7 watts

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

P(498-494.7 < Z < 498+ 494.7)

Using the Z formula :

Zscore = (x - m) / (σ/√n)

x = 498 - 494.7 = 3.3 , x = 498 + 494.7 = 992.7

Z = (3.3 - 498) / (12/√86) = -383.48

P(Z < -383.48)

Z = (992.7 - 498 ) / (12/√86) = 383.48

P(Z<383.48)

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