Consider the lot of fluorescent tubes in number 100.if 50 samples of 100 tubes are drawn from the lot with replacement find the expected number of sample will have mean life atleast 1600 hrs. Flourescent tubes are 1570 hrs with a s.d. of 150 hrs.

n=100μ=1570σ=150

Let the random variable $Y$ denote the length of life of fluorescent tubes. We determine the probability,

$p(\bar y\gt1600)=p({\bar y-\mu\over {\sigma\over \sqrt{n}}}\gt{1600-\mu\over {\sigma\over \sqrt{n}}})=p(Z\gt {1600-1570\over{150\over \sqrt{100}}})=p(Z\gt2)=1-p(Z\lt 2)=1-\phi(2)=1-0.9772=0.0228$

The expected number of samples with mean of at least 1600 is $50\times 0.0228=1.14\approx2$ samples.