In: Statistics and Probability
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
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.