Question

Suppose that prices of recently sold homes in one neighborhood have a mean of $265,000 with a standard deviation of $9300. Using chebyshev's theorem, state the range in which at least 88.9%of the data will reside

Answer 1

Answer:

Range  = (237100, 292900)

Step-by-step explanation:

Using Chebyshevs Inequality:

$P(|X - \mu | \le k \sigma )\ge 1 -\dfrac{1}{k^2}= 0.889$

$1 -\dfrac{1}{k^2}= 0.889$

$\dfrac{1}{k^2}= 1- 0.889$

$\dfrac{1}{k^2}=0.111$

$k = \sqrt{\dfrac{1}{0.111}}$

$k \simeq 3$

Thus, 88.9% of the population is within 3 standard deviation of the mean with the Range = μ  ±  kσ

where;

μ = 265000

σ = 9300

Range = 265000  ±  3(9300)

Range = 265000  ± 27900

Range =   (265000 - 27900, 265000 + 27900)

Range  = (237100, 292900)