The lifetimes of phone batteries are normally distributed with a mean of $23$ hours. They are sold in packs of $100$ to phone manufacturers. The manufacturer of the batteries aims to make batteries with more consistent lifetimes. Their target is to claim that the probability that a pack contains at least one battery with a lifetime of less than $22$ hours is below 5%. For the manufacturer to meet this target determine
(a) the minimum value of the probability of a randomly chosen battery lasting longer than $22$ hours to six decimal places
(b) the maximum expected number of batteries out of $1,000,000$ batteries with a lifetime of less than $22$ hours
(c) the maximum value of the standard deviation of the lifetime of the batteries
My solution
(a) $$1-P(X>22)^{100}<0.05$$ $$P(X>22)>0.95^{1/100}$$ $$P(X>22)>0.9994872$$
So the minimum value of $P(X>22)=0.999487$ .
(b) $$E(Y)=np=1000000(1-0.999487)=513$$
(c) We know that $P(X<22)=1-0.999487=0.00513$ So
$$\frac{22-23}{\sigma}=invNorm(0.000513,0,1)$$ $${\sigma}=2.566941$$
Are my solutions above correct?