S1 exam question: Normal distribution

This is a relatively simple question taken from the January 2005 MS/SS1B paper that I am kind of struggling with..

4) Chopped lettuce is sold in bags nominally containing 100 grams.

The weight, X grams of chopped lettuce, delivered by the machine filling the bags, may be assumed to be normally distributed with the mean M and standard deviation 4.

b) Determine the minimum value of M so that at most 2% of the bags of chopped lettuce are underweight. Give your answer to one decimal place.

c) Boxes each contain 10 bags of chopped lettuce. The mean weight of a bag of chopped lettuce in a box is denoted by (X bar).

Given that M = 108.5

i) Write down values for the mean and variance of (X bar)

ii) Determine the probability that (X bar) exceeds 110.

Any help would be truly appreciated!

Reply 1

Newton

X~N(m, (4^2))

a) P(X<100)=0.02

* Z=(1/s)(X-m)

=>Z=(1/4)(100-m)

The actual z-score is less than 0.5, there fore the z-value must lie to the left of the mean of the distribution.

P(Z<(1/4)(m-100))=0.02

By linear interpolation betweeen the values of z=1.96 and z=2.3263

=>z=2.2042

=>(1/4)(m-100)=2.2042

=>m-100=8.8168

=>m=108.8g

c) x_=108.5g=>Var(x_)=(8.5^2)=72.25

=>X~N(108.5, (8.5^2))

* Z=(1/s)(X-m)

=>Z=(1/8.5)(110-108.5)=(1.5/8.5)=0.18

P(X_>100)=1-P(X_<100)=1-P(Z<0.18)=1-0.5714=0.4286

Newton.