A meticulous gardener is interested in the length of blades of grass on his lawn. He believes that blade length X follows a normal distribution centered on 10 mm with a variance of 2 mm.
i. Find the probability that a blade of grass is between 9.5 and 11 mm long.ii. What are the standardized values of 9.5 and 11 in the context of this distribution? Using the standardized values, confirm that you can obtain the same probability you found in (i) with the standard normal density.iii. Below which value are the shortest 2.5 percent of blade lengths found?iv. Standardize your answer from (iii).

[tex]P(9.5<X<11)=P(\frac{9.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{11-\mu}{\sigma})=P(\frac{9.5-10}{1.414}<Z<\frac{11-10}{1.414})=P(-0.354<z<0.707)[/tex]

And we can find this probability with this difference:

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)=0.7602-0.3617=0.3985[/tex]

ii) The z scores for this case are:

[tex] z_1 = \frac{9.5-10}{1.414}= -0.354[/tex]

[tex] z_2 = \frac{11-10}{1.414}= 0.707[/tex]

And we can check the answer with the following excel code:

=NORM.DIST(0.707,0,1,TRUE)-NORM.DIST(-0.354,0,1,TRUE)

iii) [tex]P(X>a)=0.975[/tex] (a)

[tex]P(X<a)=0.025[/tex] (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.025 of the area on the left and 0.975 of the area on the right it's z=-1.96. On this case P(Z<-1.96)=0.025 and P(z>-1.96)=0.975

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.975[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.975[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-1.96<\frac{a-10}{1.414}[/tex]

And if we solve for a we got

[tex]a=10 -1.96*1.414=7.228[/tex]

So the value of height that separates the bottom 2.5% of data from the top 97.5% is 7.228.

iv) [tex] z = \frac{7.228-10}{1.414}= -1.96[/tex]

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

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

[tex]X \sim N(10,\sqrt{2})[/tex]

Where [tex]\mu=10[/tex] and [tex]\sigma=1.414[/tex]

Part i

For this case we want this probability:

[tex] P(9.5 < X<11)[/tex]

And we can solve this problem using the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this: