Question

masses of a particular article are normally distributed with mean 20g and standard deviation 2g. A random sample of 12 such articles is chosen. Find the probability that the total mass is greater than 230g.
Show workings.

Answer 1

The expected total weight of the 12 articles is given by:
E[W] = 20 * 12 = 240 g
The variance of the 12 samples will be the sum of the individual variances:
\[Var[W]=2^{2}\times12=48\]
The standard deviation of W will be
\[\sqrt{48}=6.928\]
We can assume that W is normally distributed with mean 240 and standard deviation 6.928
\[P(W>230)=P(Z> \frac{230-240}{6.928})=P(Z>-1.443)\]
Now you need to use a standard normal distribution table to finish.

Answer 2

sum of a normal is normal...

\[X \sim N \left( \mu, \sigma^2 \right) \Rightarrow k \cdot X \sim N\left( k\cdot \mu, k\cdot \sigma^2 \right)\]

Answer 3

Reference to a standard normal distribution table shows that:
P(Z < -1.44) = 0.075
Therefore
P(Z > -1.44) = 1.000 - 0.075 = 0.925

Answer 4

So the probability that the total mass is greater than 230g is 0.925.

Answer 5

Thanks guys! :))

Answer 6

You're welcome :)