Question

o uego 3. Metal plate thicknesses are normally distributed with a mean of 5.6mm and variance of 0.0576mm. Draw curves where relevant, 4 places past decimal: a) If 9 metal plates are averaged, what is the mean (E(X)) of the distribution of their average? What is the standard deviation? 63 1 5226 b) What is the probability that the average thickness is greater than 5.65mm ?

Answer 1

Solution:

Given: Metal plate thickness are Normally distributed with mean of $\mu=$ 5.6mm and variance of $\sigma^{2}=$ 0.0576 mm.

then standard deviation =

0 = VOP= V0.0576 = 0.24

Part a) Sample size = n = 9

Mean (E(X)) of the distribution of sample average is:

Vi=

Mi = 5.6

standard deviation of sample means is:

0i = n

0.24 0i =

0.24 Oi=

0 = 0.08

Part b) Find:

PCI > 5.65) = .......

Find z score for

= 5.65

1 - Hi 2=

5.65 – 5.6 0.05 0.08 = = 0.625 = 0.63

Thus we get:

PT > 5.65) = P(Z > 0.63)

PT> 5.65) = 1- P(Z < 0.63)

Look in z table for z = 0.6 and 0.03 and find corresponding area.

08 .09 「 .00 ,01 ,02 ,03 .04 10.0 1,5000 1.5040 .5080 1.5120 5160 10.1 5398 15438 5478 57.5557 10.2 .57935832 .5871 54 10 .5948 0.3 .6179 .6217 ,6255 .6193 .6331 10.4 ,6554 .6591 .6628 1.6464 .6700 0.5 ,6915 .6950 .6985 7019 7054 10.6 - 72947324-7357 .7389 .05 5199 5596 1.5987 .6368 .6736 7088 7422 .06 .07 .08 .5239 5279 .5319 .5636 .5675 ,5714 .6026 ,6064 .6103 .6406 .6443 .6480 .6772 ,6808 ,684 7123 ,7157 7190 ,7454_7486 7517 1.5359 15753 ,6141 .6517 .6879 ,7224 ,7549 |

P( Z< 0.63)= 0.7357

Thus

PT> 5.65) = 1- P(Z < 0.63)

PT> 5.65) = 1-0.7357

PT> 5.65) = 0.2643

Pl> 5.65 ) = 0.2643 Mean =5.6mm i =5.65