Question

In: Statistics and Probability

A packet contains six biscuits, each of which is individually wrapped. The mass of a biscuit...

A packet contains six biscuits, each of which is individually wrapped. The mass of a biscuit can be taken to be normally distributed with mean 70 g and standard deviation 4 g. The mass of the individual wrapping of a biscuit is normally distributed with mean 10 g and standard deviation 1 g. The mass of the outer packaging is normally distributed with mean 30 g and standard deviation 3 g. Assuming that the masses of the biscuits, wrappings and packaging are independent, calculate the probability that the total mass of a randomly chosen packet and its contents lies between 500 g and 520 g

Solutions

Expert Solution

Since a packet contains : 6 biscuits each distributed Normally with mean=70 g and sd=4 g denote it by X

6 individual wrappings distributed Normally with mean=10 g and sd=1 g denote it by Y

1 outer packaging Normal with mean =30 g and sd=3 g. denote it by Z

Each of these three are independent of each other.

In Normal theory, we know that if X, Y and Z are independent, we have

E(aX+bY+cZ)=a*E(X)+b*E(Y)+c*E(Z) and

If we take a=6, b=6 and c=1 we shall get the mean of the total mass of the packet as

6*70+6*10+1*30=510 g

The variance of the total mass is

  

Therefore the mass of the packet is also Normal with mean=510 gand variance=621 g or sd=24.91g.

Now the probability that the total mass of a randomly chosen packet and its contents lies between 500 g and 520 g

ie .

When P=500,

P=520,

  

  

Therefore, the probability that the total mass of a randomly chosen packet and its contents lies between 500 g and 520 g = 0.3118.


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