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.

Step-by-step explanation:

mean x =

Variance

standard deviation

n = 400

confidence :

From standard normal distribution table,

Margin of error,

Lower limit: x - E

Upper limit: x + E

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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 level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which 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

Then

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.

for any of you who need the translation here it is

step-by-step explanation:

find the angular size of the equatorial radius of the earth, if the horizontal parallax of the sun