Question

The distances male long jumpers for State College jump are approximately normal with a mean of 263 inches and a standard deviation of 14 inches. Suppose a male long jumper's jump is ranked in the 75th percentile. How long was his jump?

Answer 1

Answer:

His jump was of 272.45 inches

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 263, \sigma = 14$

75th percentile

X when Z has a pvalue of 0.75. So X when Z = 0.675

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 263}{14}$

$X - 263 = 14*0.675$

$X = 272.45$

His jump was of 272.45 inches

Answer 2

Answer:

$z=0.674$

And if we solve for a we got

$a=263 +0.674*14=272.436$

So the value of height that separates the bottom 75% of data from the top 25% is 272.436.  

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 distances of a population, and for this case we know the distribution for X is given by:

$X \sim N(263,14)$  

Where $\mu=263$ and $\sigma=14$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.25$   (a)

$P(X$   (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 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

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

$P(X$  

$P(z$

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

$z=0.674$

And if we solve for a we got

$a=263 +0.674*14=272.436$

So the value of height that separates the bottom 75% of data from the top 25% is 272.436.