Question

If x is a normal variable with the mean u=5 and variance (sigma square)=16, what is the probability that x is less than or equal to 6?

Answer 1

Answer on Question #68050 – Math – Statistics and Probability

Question

If $X$ is a normal variable with the mean $\mu = 5$ and variance $(\sigma^2) = 16$, what is the probability that $X$ is less than or equal to 6?

Solution

First of all, note that if $\sigma^2 = 16$ then $\sigma = 4$. Further, if $X$ is a normal variable then

$$Y = \frac{X - \mu}{\sigma} = \frac{X - 5}{4}$$

is a normal variable with mean 0 and variance (and standard deviation) 1

(see https://en.wikipedia.org/wiki/Normal_distribution). The required probability is

$$P\{X \leq 6\} = P\left\{\frac{X - 5}{4} \leq \frac{6 - 5}{4}\right\} = P\{Y \leq 0.25\}.$$

Due to the symmetry of the distribution of $P\{Y \leq 0\} = 0.5$. Then $P\{Y \leq 0.25\} = P\{Y \leq 0\} + P\{0 < Y \leq 0.25\} = 0.5 + P\{0 < Y \leq 0.25\}$. The last probability we can find from the Standard Normal Distribution table (see https://www.mathsisfun.com/data/standard-normal-distribution-table.html): $P\{0 < Y \leq 0.25\} = 0.0987$. Then $P\{Y \leq 0.25\} = 0.5 + 0.0987 = 0.5987$.

Answer: 0.5987.

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