Question

7. Suppose that the average weekly earnings for employees in general automotive repair shops is $450, and that the standard deviation for the weekly earnings for such employees is SSO. A sample of 100 such employees is selected at random. a) Find the mean and standard deviation of the sampling distribution of the average weekly earnings in the sample. (b) Find probability that the mean of the sample is less than $445. (e) Find the probability that the mean of the sample is between $445 and $455. (d) Find the probability that the mean of the sample is greater than $460.

Answer 1

♡ Here, we have mean (M) = 450 5 Jandard deviation O) 50 sample size in 2 100 (9) mean I = H: 4.50 stand and deviation cannot shoot-01

(b) PL 3 445) - Pl Andy 3375 445- 450 per la s 4452 450) ہ = Plz <-1) = 0.158855 Ibrom standard normal calculators - 10.1987) Round to 4 deeimal places ☺) AL 445535455)= P(445-450 < P-4 <455-49 =pl-1226) = PIZSb - PLZS4) = 0.841345 – 0.158655 2 0.68 269 - 10.6827] Round to 4 decimal places (d) Plx> 470) = P1 R , 470-450) x Ĉ PL Z ,4) a 1- P1Z34) -1- 0.999968 (from standard =10.000032 normal calcwater)

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