The random variable $X \sim N(\mu, 5^2)$ and $P(X < 23) = 0.9192$ (a) Find the value of $\mu$ - Edexcel - A-Level Maths Statistics - Question 2 - 2011 - Paper 2

Since we have found $\mu = 16$, we need to calculate:

$P(16 < X < 23)$

This is equivalent to:

$P(X < 23) - P(X < \mu) = P(X < 23) - P(X < 16)$

Using the properties of the normal distribution:

1. We know that $P(X < 23) = 0.9192$.
2. From the standard normal distribution, we can find that $P(X < 16)$ corresponds to a Z-value of -1.0.

Thus, $P(X < 16) = 0.1587$. Therefore:

$P(16 < X < 23) = 0.9192 - 0.1587 = 0.7605.$

In conclusion, the answer is:

$P(16 < X < 23) = 0.7605$.