Question

x is normally distributed with a mean of 100 and a standard deviation of 20. what is the probability that x is greater than 145

Answer 1

The probability density function for normal (gaussian) distribution is

$$\frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{{{{(x - \mu )}^2}}}{{2{\sigma ^2}}}}}$$

In this case mean $\mu = 100$ and standard deviation $\sigma = 20$ .

1)Use standard normal variable Z

$$Z = \frac{{X - \mu }}{\sigma } \Rightarrow X = \sigma Z + \mu$$

so the probability that the random variable X is greater than 145 can be represented in standard form

$$P(X > 145) = P(20Z + 100 > 145) = P(Z > 2.25)$$

Now we can use standard normal distribution table (Z-table) (I get it from z-table.com)

It represents the probability that $Z < a$, so we need to use the normalized property (full probability is equal to 1), so $P(Z > a) = 1 - P(Z < a)$. In our case from table $P(Z < 2.25) \approx 0.9878$ so

$$P(X > 145) = P(Z > 2.25) = 1 - P(Z < 2.25) = 1 - 0.9878 = 0.0122$$

The answer is 0.0122.

Answer: $P(X > 145) \approx 0.0122$