Normal distribution

3. From a set of 1000 observations known to be normally distributed, the mean is 534 cm and SD is 13.5 cm. How many observations are likely to exceed 561 cm? How many will be between 520.5 cm and 547.5 cm? Between what limits will the middle 50% of the observations lie?

I'll show you how to do the first part.

Let X represent the length of one observation.

X \sim N(534, 13.5^2)

The probability that you get one observation greater than 561 cm si given by:

P(X > 561) = P(z > \frac{561 - 534}{13.5})

Find the z score and look up your table for the corresponding probability.

You should get

P(z > 2) = 1- 0.9772 = 0.0228

Hence, the expected number of observations above 561 will be:

Expectation = np = 1000 \times 0.0228 = 22.8\ observations