Suppose the weights of male police are normally distributed and the 6.68% are under 130 lbs. in weight, and 77.45% are
between 130 and 180 lbs. Find the mean and standard deviation of the distribution.
You can play around with Z-table stuff here:
http://davidmlane.com/hyperstat/z_table.html
To find the mean and standard deviation of the distribution, we can use the properties of the normal distribution and the given information.
First, let's define the variables:
- Let μ be the mean of the distribution.
- Let σ be the standard deviation of the distribution.
Given:
- 6.68% of males are under 130 lbs, which means the area to the left of 130 lbs is 0.0668.
- 77.45% of males are between 130 and 180 lbs, which means the area between 130 and 180 lbs is 0.7745.
We need to find the values of μ and σ.
To find the mean (μ):
Since the distribution is symmetric, the mean (μ) will be the middle point between 130 lbs and 180 lbs. Therefore:
μ = (130 + 180) / 2
= 310 / 2
= 155 lbs
Thus, the mean (μ) of the distribution is 155 lbs.
To find the standard deviation (σ):
We know that 77.45% of the males fall within one standard deviation of the mean in a normal distribution. Moreover, since the distribution is symmetric, half of this percentage will be between the mean and 180 lbs.
Therefore, the area between the mean and 180 lbs is: (77.45% / 2) + (50% - 6.68%)
= (0.7745 / 2) + (0.50 - 0.0668)
= 0.38725 + 0.4332
= 0.82045
Using a standard normal table or a statistical calculator, we can find the z-score associated with an area of 0.82045. The z-score is 0.902.
To convert this z-score to an actual weight value, we can use the formula:
z = (x - μ) / σ
Substituting the known values:
0.902 = (180 - 155) / σ
25 = 0.902σ
σ ≈ 27.704
Thus, the standard deviation (σ) of the distribution is approximately 27.704 lbs.
In summary, the mean (μ) of the distribution is 155 lbs, and the standard deviation (σ) is approximately 27.704 lbs.