Question

The average heights of a random sample of 400 people from a city is 1.75 m. It is known that the heights of the population are random variables that follow a normal distribution with a variance of 0.16.
Determine the interval of 95% confidence for the average heights of the population.

Answer 1

Answer:

The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So

$\sigma = \sqrt{0.16} = 0.4$

Then

$M = 1.96*\frac{0.4}{\sqrt{400}} = 0.0392$

The lower end of the interval is the mean subtracted by M. So it is 1.75 - 0.0392 = 1.7108m

The upper end of the interval is the mean added to M. So it is 1.75 + 0.0392 = 1.7892m

The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.