How is this answer achieved?

Question

If X is a normal random variable with mean 11, and if the probability that X is less than 12.10 is .72, then what is the standard deviation of X?

Answer 1

For any normal random variable X with mean μ and standard deviation σ, X ~ Normal(μ, σ), { note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal(μ, σ²). Most software denotes the normal with just the standard deviation.}

you can translate into standard normal units by:
Z = (X - μ) / σ

where Z ~ Normal(μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.

we have in this problem

P( X< 12.10) = 0.72

we know that

P( Z < 0.5828415) = 0.72

P( Z < (12.10 - 11) / σ = 0.5828415 ) = 0.72

(12.10 - 11) / σ = 0.5828415

σ = (12.10 - 11) / 0.5828415

σ = 1.887306

Answer 2

The cumulative distribution function F(x) is defined as:

F(x) = P( X <= x)

The CDF of a normal random variable is completely specified by mean and standard deviation. That is,

F(x) = 0.5*(1 + erf( (x-mu)/(sigma*sqrt(2)) )

where "erf" is the error function, mu is the mean, and sigma is the standard deviation.

In your case, you have been given:

F(12.10) = P(X <= 12.10)

So, you need to solve

0.72 = 0.5*(1 + erf( (12.10-11)/(sigma*sqrt(2)) )

for sigma. Because of the error function, this problem should be solved numerically (e.g., in MATLAB, Excel, etc.).

Equivalently, you can use a table of CDF values for a normal random variable with 0 mean and unity standard deviation. Find the "z" value corresponding to a probability of 0.72. Then convert to your distribution with:

z = (x - mu)/sigma

That is, solve for sigma with:

sigma = (12.10 - 11)/z

where "z" is the point you took from the table (corresponding to a probability of 0.72). You can find some sample tables in the sources in this answer. For a probability of 0.7224, you get a z-score of 0.59. So, you'd get a sigma of

sigma = (12.10 - 11)/0.59 = 1.86

For a better answer, find a table with more resolution (or use MATLAB, Excel, SAS, SPSS, etc.).

To verify my answer, use the Java applet (the last source in my list below). You should see that for a mean of 11 and standard deviation of 1.86, the probability that the random variable is less than 12.1 is 0.722873.

Answer 3

The usual way. Go to your friendly z-table and find the normalized standard deviation for which [12.1-11]/11 would be equal to a cumulative prob dist of 0.72. This is about 1.05 or so, so the standard deviation would be 1.1/1.05 or about 1.05.