Question

a populatpion of 1,000 students has an average weekly allowance of =350 and standard deviation of =56.13 .what is the probability that a random sample of size =30 will have an average weekly allowance between 335 and 360?

Answer 1

We have a normal distribution, $\mu=350, \sigma=56.13.$

Let's convert it to the standard normal distribution, $z=\cfrac{\bar{x}-\mu}{\sigma/\sqrt{n}};$

$z_1=\cfrac{335-350}{56.13/\sqrt{30}}=-1.46, \\z_2=\cfrac{360-350}{56.13/\sqrt{30}}=0.98,$

$P(335<X<360)=P(-1.46<Z<0.98)=\\=P(Z<0.98)-P(Z<-1.46)=$

$=0.83646-0.07215=0.76431$ ​(from z-table).