For a normal distribution with µ = 80 and σ = 10 the score that separates the bottom 10% of the distribution from the rest is 67.2.

Question

Please show the working too. I am trying to understand this question.

Answer 1

z score for 80% confidence interval is 1.282 You can find z tables or find the score with a calculator or online

the empirical rule is 68-95-99 or roughly 68% within one standard deviation of the mean, 95% within two, and 99% within 3.

80% be somewhere between 1 and 2 standard deviations from the mean. 68+95=163 163/2=81.5 The 80% confidence interval should be a little closer to 1 than 2 standard deviations, and less than 1.5.

You need 80 because that means 10% will be above that interval and 10% below that interval. You want the bottom 10% which is 1.282 standard deviations below 80.

67.2 is 12.8 less than 80. 12.8 divided by 10 = 1.28, the z-score for an 80% confidence interval.

Take the difference between 67.2 and the mean of 80 and divide by the standard deviation, that gives you the z-score for an 80% confidence interval.

Answer 2

First, let's find the Z score that corresponds to the 10th percentile on a normal distribution.

Using this table: https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf

We can see that Z = -1.28 corresponds to the 10th percentile of the distribution.

The formula for a Z statistic is Z = (x - mu)/sigma where mu = population mean and sigma = population standard deviation. x is the value that corresponds to the given z statistic. For example, if x=mu, then Z = (mu-mu)/sigma = 0 and the corresponding z statistic is 0.

Thus:

-1.28 = (x - mu)/sigma

-1.28 = (x-80)/10

-12.8 = x-80

x = 67.2

So this is how you get the answer of 67.2.