The resistance of a certain type of electrical resistors has normal distribution with mean of 45 ohms. A random sample of 9 such resistors is selected, and the variance is computed to be 4. What is the probability that the average resistance will exceed 47 ohms?

0.00135

Step-by-step explanation:

When given a random sample of numbers:

z-score is is z = (x-μ)/σ/√n

where x is the raw score

μ is the population mean

σ is the population standard deviation.

Standard deviation = √variance = √4 = 2

Mean = 45 ohms

A random sample of 9 such resistors

z = 47 - 45/2 / √9

z = 2/2/3

z = 3

P-value from Z-Table:

P(x<47) = 0.99865

P(x>47) = 1 - P(x<47) = 0.0013499

Approximately ≈ 0.00135

The probability that the average resistance will exceed 47 ohms is 0.00135

idk bring the function back   im not smart

step-by-step explanation:

answer: 140

step-by-step explanation:

so just add all then it will be your answer but i am not pretty sur eif it is but hope this i am here to people