What is the probability that the 36 tires will have an
average of less than 16,000 miles until the tires begin
to wear out
We have that: "X\\sim N(\\mu, \\sigma^2\/n)"
"\\mu=16800"
"\\sigma=3300"
"n=36"
"P(\\bar X<16000)=P(Z<\\dfrac{X-\\mu}{\\sigma\/\\sqrt{n}})""=P(Z<\\dfrac{16000-16800}{3300\/\\sqrt{36}})=P(Z<-1.454545)""\\approx0.0729"
Comments
Leave a comment