Question

a) if you have scores and you dont know the shape of their distribution, find the minimum proportion of scores that fall within 2.8 standard deviations on both sides of the mean? round to two decimal places.
b) if you have scores and you dont know the shape of their distribution, how many standard deviations on both sides of the mean do we need to go to have 41 percent of the scores? round to one decimal place.
hint: in part b you will need to solve for k in the chebyshevs equation.

Explanation:

Step1: Recall Chebyshev's theorem formula

The formula for Chebyshev's theorem is $P(|X - \mu|\leq k\sigma)\geq1-\frac{1}{k^{2}}$, where $k$ is the number of standard - deviations from the mean, $\mu$ is the mean, $\sigma$ is the standard deviation, and $P(|X - \mu|\leq k\sigma)$ is the proportion of data within $k$ standard deviations of the mean.

Step2: Solve part a

Given $k = 2.8$. Substitute $k$ into the Chebyshev's formula:
$P(|X - \mu|\leq2.8\sigma)\geq1-\frac{1}{2.8^{2}}$.
First, calculate $\frac{1}{2.8^{2}}=\frac{1}{7.84}\approx0.1276$.
Then, $1-\frac{1}{7.84}=1 - 0.1276 = 0.8724\approx0.87$.

Step3: Solve part b

We know that $P(|X - \mu|\leq k\sigma)=0.41$. Using Chebyshev's formula $1-\frac{1}{k^{2}} = 0.41$.
Rearrange the formula to solve for $k$:
$\frac{1}{k^{2}}=1 - 0.41=0.59$.
Then, $k^{2}=\frac{1}{0.59}\approx1.6949$.
Take the square - root of both sides: $k=\sqrt{\frac{1}{0.59}}\approx1.3$.

Answer:

a) 0.87
b) 1.3