Question Type 2: Finding the interval for which a specific percentage of data is concentrated Exercises

A dataset has mean $\mu = 60$ and you know that 99% of the values lie between 30 and 90. Use Chebyshev's theorem to find the variance $\sigma^2$.

Suppose a variable has mean $\mu=100$. Using Chebyshev’s inequality, find the maximum possible variance $\sigma^2$ such that at least $80\%$ of the observations lie between 70 and 130.

Assume a distribution with mean $\mu = 0$ and variance $\sigma^2 = 25$. By Chebyshev's theorem, determine the symmetric interval about the mean that must contain at least 95% of the data.

A distribution has $\mu = 0$ and $\sigma^2 = 4$. Use Chebyshev's theorem to find the symmetric interval about the mean that contains at least 88.9% of the data.

Given a random variable with mean $\mu = 8$ and it is known that 95% of the data lies between 6.5 and 9.5, use Chebyshev’s inequality to determine the minimum possible value of the variance $\sigma^2$.

A dataset has mean $\mu = 50$ and variance $\sigma^2 = 16$. Use Chebyshev's theorem to find the interval around the mean that contains at least $90\%$ of the data.

For a normally distributed variable with mean $\mu=0$, $95\%$ of the data lies between $-9.8$ and $9.8$. Determine the variance $\sigma^2$.

A random variable with mean $\mu = 200$ has $75\%$ of its data between $170$ and $230$. Find $\sigma^2$ using Chebyshev's theorem.

A variable is normally distributed with mean $\mu=50$ and 99.7% of values between 35 and 65. Find the variance $\sigma^2$.

Given a random variable with mean $\mu = 20$ and variance $\sigma^2 = 9$, use Chebyshev's theorem to determine the interval centered at the mean that must contain at least 80% of the data.

A random variable has mean $\mu = 100$ and variance $\sigma^2 = 36$. Use Chebyshev's theorem to find the smallest interval centered at 100 that contains at least 96% of the data.

For a normally distributed variable with mean $\mu = 100$, 95% of the data falls between 85 and 115. Determine the variance $\sigma^2$.

Assume a normal distribution with mean $\mu = 0$ and you observe that $68\%$ of data falls within $\pm5$. Find the variance $\sigma^2$.

Given a mean $\mu = 0$ and variance $\sigma^2 = 16$, determine the length of the smallest symmetric interval about the mean that contains at least 75% of the data using Chebyshev's theorem.