Standard Deviation: The masses of sea lions on an island are normally distributed with mean x and

standard deviation s. 10% of them have mass greater than 900 kg and 15% of them have mass less than 500 kg. Find the mean x and the standard deviation s.

P(X ≥ 900) = 0.1

P(X ≤ 500) = 0.15

P((X - μx)/σx ≥ (900 - μx)/σx) = 0.1

P(Z ≥ z) = 0.1

z = 1.28

1.28 = (900 - μx)/σx

1.28*σx = 900 - μx

P((X - μx)/σx ≤ (500 - μx)/σx) = 0.15

P(Z ≤ - z) = 0.15

z = - 1.045

-1.045*σx = 500 - μx, multiply by 1.28
1.28*σx = 900 - μx, multiply by 1.045
-----------------------------

- 1.3376*σx = 640 - 1.28*μx
...1.3376*σx = 940.5 - 1.045*μx
--------------------------------------…
0 = 1580.5 - 2.325*μx

μx = - 1580.5/( - 2.325)

μx = 679.8

1.3376*σx = 940.5 - 1.045*μx

σx = (940.5 - 1.045*679.8)/1.3376

σx = 172

There should be a table in your statistics book that says how many standard deviations above the mean 900 kg is, based on the 90% probability that a sea lion weighs less than 900 kg. This is sometimes called a z-score. This number is +1.28.

In this same table look for the z-score corresponding to 15% of the sea lions weighing less than 500 kg.This value is -1.04.

900 kg minus 500 kg corresponds to 1.28 - (-1.04) = 2.32 standard deviations.

Then one standard deviation s equals 400 kg/2.32.

Now that you know s, you can compute x by solving the following equation.

x + 1.28s = 900.