Answer to Question #299629 in Statistics and Probability for Sinbad

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

Let $X=$ an average weekly allowance: $X\sim N(\mu, \sigma^2/n).$

Given $\mu=350, \sigma=56.13, n=30$

$$P(335<X<360)=P(X<360)-P(X\le335)$$

$$=P(Z<\dfrac{360-350}{56.13/\sqrt{30}})-P(Z\le\dfrac{335-350}{56.13/\sqrt{30}})$$

$$\approx P(Z<0.975811)-P(Z\le-1.463716)$$

$$\approx 0.83542086-0.07163577$$

$$\approx 0.7638$$