The average price of a television on a certain Web site is $830. Assume the pric

Question

The average price of a television on a certain Web site is $830. Assume the price of these televisions follows the normal distribution with a standard deviation of $190. Complete parts a through d below a. What is the probability that a randomly selected television from the site sells for less than $700? (Round to tour decimal places as needed) b. What Is the probability that a randomly selected television from the site sells for between $400 and $500? (Round to tour decimal places as needed) c. What is the probability that a randomly selected television from the site sells for between $900 and $1.000? (Round to four decimal places as needed) d. There are 17 televisions on the she How many televisions on the site. How many televisions are within a $800 budget? There are televisions on the site that are within a $800 budge (Round down to the nearest whole number.)

Explanation / Answer

Answer to the question below are:

a. P(X<700) = P(Z< 700-830 / 190) = P(Z<-.68) = 1-.2483 = .7517

b. P(400<X<500) = P(400-830/190 <Z< 500-830/190) = P(-2.26<Z<-1.73)= .0418-.0119 = .0299

c. P(900<X<1000) = P(.895<Z<.368) = .8159-.6443 = .1716

d. n = 17

Within the $800 budget, we have: P(X<800) = P(Z<(800-830) / (190/sqrt(17)) = P(Z<-.65) = .2578

So, 17*.2578 = 4.38 people or 4 people

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