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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!

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!

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.

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.

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can someone please explain what principle domain is and why the answer is a not c?Maths

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