Question

Assume that the readings on the thermometers are normally distributed with a mean of 0° and standard deviation of 1.00°C. A thermometer is randomly selected and tested.
Draw a sketch and find the temperature reading corresponding to P97​, the 97th percentile. This is the temperature reading separating the bottom 97% from the top 3%.

Answer 1

Answer:

So the 97 percentile for this case should be 1.88 degrees.

Explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the temperatures of a population, and for this case we know the distribution for X is given by:  

$X \sim N(0,1)$  

Where $\mu=0$ and $\sigma=1$.  

And we want the top 97 percentile for the distribution, so we need a value a such that:

$P(X>a)=0.03$ or $P(X$

We need on the right tail of the distribution a value a that gives to us 97% of the area below and 3% of the area above. Both conditions are equivalent.

Let's use the condition $P(X$, the best way to solve this problem is using the z score with the following formula:

$z=\frac{x-\mu}{\sigma}$

So we need a value from the normal standard distribution that accumulates 0.97 of the area on the left and 0.03 on the right. This value on this case is 1.88 and we can founded with the following code in excel:

"=NORM.INV(0.97,0,1)"

The figure attached gives an approximation of the value and we can see that is near 1.88.

If we apply the z score formula to our case we have this:

$P(X$

So then based on the equalities we have this:

$\frac{a-0}{1}=1.88$

And if we solve for a we got:

$a=(1*1.88) +0=1.88$

So the 97 percentile for this case should be 1.88 degrees.

The second figure attached illustrate the solution for this case.