Answer:
a. 0.71
b. 0.9863
Step-by-step explanation:
a. The mean of the distribution is given as $403,000 and the standard deviation is $278,000.
-To estimate the probability that a randomly selected house has a value less than $500,000:
[tex]P(X<500,000)=P(X=0)+P(X=500)\\\\=0.34+0.37\\\\=0.71[/tex]
Thus, the probability that a randomly selected house has a value less than $500,000 is 0.71
b. -since 40 is larger than or equal to 30, we assume normal distribution.
-The probability can therefore be calculated as:
[tex]P(\bar X)=P(z<\frac{\bar X-\mu}{\sigma/\sqrt{n}})\\\\=P(z<\frac{500-403}{278/\sqrt{40}})\\\\=P(z<2.2068)\\\\=0.986336\\\\\approx 0.9863[/tex]
Hence, the probability that the mean value of the 40 houses is less than $500,000 is 0.9863
