Question

1. Suppose that a factory makes machine parts, and the size of any of the machine parts X is normally distributed with mean = 8.3 inches and standard deviation o = 1.4 inches. a) What is the probability that a single random machine part is between 8.2 and 8.5 inches? b) Assume that the different parts made by the factory are independent, and all with the same normal distribution given above (we say they are 'i.i.d.'). If you randomly choose seven of these parts and compute the average size of the seven, X, find the probability that X is between 8.2 and 8.5 inches. c) What is the probability that, if you add up the sizes of the seven parts, the total sum S of the seven parts has a total value between 56 and 62 inches?

Answer 1

a)

Let X is a random variable shows the given distribution. Given that: A = 8.3, J =1.4

The Z-score for X = 8.2 is: X-1 = (8.2 -8.3)/1.4 = -0.07 The Z-score for X = 8.5 is: X- = o = (8.5 - 8.3)/1.4 = 0.14 Using z table, the probability that X is between 8.2 and 8.5 is: P( 8.2 <= X<= 8.5 ) = P( -0.07 <= Z <= 0.14) = P(Z <= 0.14 ) - P(Z <= -0.07 ) = 0.5557 -0.4721 = 0.0836 Answer: 0.0836 Excel can be used to find the probabilites is =ROUND(NORMSDIST(0.14) - NORMSDIST(-0.07), 4).

b)

Since it is given that population is normally distributed so using CLT the sampling distribution of sample mean will be approximately normal with mean and SD as follows: W = = 8.3 x Jn n = 1.4/sqrt(7) = 0.5292

The Z-score for =8.2 is: Ille ZE 6% = (8.2 -8.3)/0.5292 = -0.19 The z-score for i=8.5 is: K-llo ZE og = (8.5 - 8.3)/0.5292 = 0.38 Using z table, the probability that I is between 8.2 and 8.5 is: P(8.2<= <= 8.5 ) = P(-0.19<= z<= 0.38) = P(Z <= 0.38 ) - Plz<= -0.19 ) = 0.648 -0.4247 = 0.2233 Answer: 0.2233 Excel can be used to find the probabilites is =ROUND(NORMSDIST(0.38) - NORMSDIST(-0.19), 4).

(c)

Since it is given that population is normally distributed so using CLT the sampling distribution of sample sum will be approximately normal with mean and SD as follows: Is = \l = 58.1 Os = 0.vn = 1.4/sqrt(7) = 3.7041

The Z-score for S =56 is: S-lls Os = (56 - 8.3)/3.7041 = -0.57 The Z-score for S=62 is: 2= S- με os = (62 - 8.3)/3.7041 = 1.05 Using z table, the probability that S is between 56 and 62 is: P( 56 <=S <= 62 ) = P(-0.57 <= 2 <= 1.05 ) = P(z<= 1.05 ) - Plz<=-0.57 ) = 0.8531 - 0.2843 = 0.5688 Answer: 0.5688 Excel can be used to find the probabilites is =ROUND(NORMSDIST(1.05) - NORMSDIST(-0.57), 4).

-----------------------------------------------------