Question

A probability distribution has a mean of 80 and a standard

deviation of 4. Use Chebyshev's inequality to find the minimum
probability that an outcome is between 48 and 112.

Answer 1

Chebyshev's inequality states that at least 95 percent of the data is within 4.5 standard deviations of the mean. To find the minimum probability that an outcome is between 48 and 112 with a mean of 80 and a standard deviation of 4, calculate the range of values within 4.5 standard deviations of the mean.

Step-by-step explanation:

Chebyshev's inequality states that for any data set, no matter what the distribution of the data is:

1. At least 75 percent of the data is within two standard deviations of the mean.
2. At least 89 percent of the data is within three standard deviations of the mean.
3. At least 95 percent of the data is within 4.5 standard deviations of the mean.

Therefore, in your case, with a mean of 80 and a standard deviation of 4, you can use Chebyshev's inequality to find the minimum probability that an outcome is between 48 and 112.

To find the minimum probability, you need to find the range of values within 4.5 standard deviations of the mean (80). Calculate the upper limit by adding 4.5 standard deviations to the mean: 80 + (4.5 * 4) = 98. Then calculate the lower limit by subtracting 4.5 standard deviations from the mean: 80 - (4.5 * 4) = 62. Therefore, the minimum probability that an outcome is between 48 and 112 is the probability that an outcome is between 62 and 98.