Answer to Question #152044 in Statistics and Probability for JG

Question

Based on a study conducted on 10,000 houses in a countryside area, a realtor found the houses in this area is normally distributed with a mean of 2100 square feet, and a standard deviation of 600 square feet.
a) If we randomly select 6 houses, what is the probability that the average of these houses is above 2000 square feet?
b) Approximately how many houses are above 1500 square feet?
c) A house is selected at random, what is the probability that it is between 1700 to 2800 square feet?
d) Knowing a house is exactly 10% above the first quartile, what is its size?

If we examine the histogram, and found the size of these houses actually follows a left skewed distribution. Use this information to answer e) and f).
e) At least how many percent of the houses in this area are between 1320 to 2880 square feet.
f) At least 7700 houses in this area are between what ranges in square feet.

Answer 1

μ = 2100

σ = 600

$Z = \frac{\bar{X}-μ}{\frac{σ}{\sqrt{n}}} \\ (a)\; P(\bar{X}>2000) = P(Z> \frac{2000 – 2100}{\frac{600}{\sqrt{6}}}) \\ = P(Z>-0.41) \\ = 0.6591 \\ (b) \;Z = \frac{X-μ}{σ} \\ P(X>1500) = P(Z>\frac{1500-2100}{600}) \\ = P(Z>-2.45) \\ = 0.9929$

Expected number of boxes $= 10000 \times 0.9929$

$= 9929 \\ (c) \;P(1700<X<2800) = P(\frac{1700-2100}{600}<Z< \frac{2800-2100}{600}) \\ = P(-1.63 <Z<2.86) \\ = 0.9463 \\$

(d) P = 25 % + 10 % = 35 %

Z-score = -0.383

$Z = \frac{X-μ}{σ} \\ -0.383 = \frac{X-2100}{600} \\ X = 1870$

(e)(f) no histogram

(e)(f) no histogram