Question

The average home in Indiana in 2015 sold for on average $105,000, with a variance of 900,000,000 (or a standard deviation of 30,000). i.e. ~??(105,000 ,900,000,000).
a. What is the probability that a home chosen at random is sold at a price less than $95,000?
b. What is the probability that a home chosen at random is sold at a price less than $115,000?
c. What is the probability that a home chosen at random is between $95,000 and $115,000?

Answer 1

Solution :

Given that ,

a.

P(x < 95000) = P[(x - ) / < (95000 - 105000) / 30000]

= P(z < -0.3333)

= 0.3695

Probability = 0.3695

b.

P(x < 115000) = P[(x - ) / < (115000 - 105000) / 30000]

= P(z < 0.3333)

= 0.6305

Probability = 0.6305

c.

P(95000 < x < 115000) = P[(95000 - 105000)/ 30000) < (x - ) /  < (115000 - 105000) / 30000) ]

= P(-0.3333 < z < 0.3333)

= P(z < 0.3333) - P(z < -0.3333)

= 0.6305 - 0.3695

= 0.261

Probability = 0.261