The random variable X has a normal distribution with mean mu and standard deviation sigma. The mean and standard deviation are also related by the equation 2sigma = 5mu, where mu eq 0. Calculate P(X >

Given $2\sigma = 5\mu$, we can write:

The Z-score formula is $Z = \frac{X - \mu}{\sigma}$.
Substitute $X = 5\mu$ and $\sigma = 2.5\mu$:

Using a standard normal distribution table or calculator, $P(Z \le 1.6) \approx 0.9452$.
Therefore, $P(Z > 1.6) = 1 - P(Z \le 1.6)$:

The correct option is A.
The probability $P(X > 5\mu)$ is $\boxed{\text{A. } \textbf{0.0548}}$.

19. In a Durban suburb, 10% of houses have solar geyser. 16 houses are chosen at random. Find the probability that fewer than 4 houses have solar geyser.

This is a binomial distribution problem, where $n$ is the number of trials and $p$ is the probability of success.
Number of houses chosen, $n = 16$.
Probability of a house having a solar geyser, $p = 0.10$.
Probability of a house not having a solar geyser, $1-p = 0.90$.
We need to find the probability that fewer than 4 houses have a solar geyser, which means $P(X < 4)$ or $P(X \le 3)$.
This is $P(X=0) + P(X=1) + P(X=2) + P(X=3)$.

For $k=0$:

For $k=1$:

For $k=2$:

For $k=3$:

Rounding to four decimal places, $P(X < 4) \approx 0.931