Find the corresponding z-score to the given value and determine whether the value is unusual.
A weight of 225 pounds among a population having a mean weight of 161 pounds and a standard deviation of 23.0 pounds.
Holy smoke!
By the looks of things, Joe/Johnny/Josh/John needs to stop now and re-take this entire course from the beginning. Thirteen posts -- and not a single thought of your own??
=(
you can play around with Z table stuff at
http://davidmlane.com/hyperstat/z_table.html
To find the corresponding z-score for a value, we need to use the formula:
z = (x - μ) / σ
where:
- x is the given value
- μ is the mean of the population
- σ is the standard deviation of the population
In this case, the given value (x) is the weight of 225 pounds, the mean (μ) is 161 pounds, and the standard deviation (σ) is 23.0 pounds.
Let's plug these values into the formula:
z = (225 - 161) / 23.0
z = 64 / 23.0
z ≈ 2.783
The corresponding z-score is approximately 2.783.
To determine if the value is unusual, we can compare the z-score to the standard normal distribution.
In a standard normal distribution, which has a mean of 0 and a standard deviation of 1, z-scores greater than 2 or less than -2 are considered unusual (or outliers). These values typically fall outside of 95% of the data.
Since the z-score of 2.783 is greater than 2, we can conclude that the weight of 225 pounds is unusual or atypical in this population, assuming that the distribution of weights follows a normal distribution.