z-score edits

13-foundations-mathematical.qmd

@@ -726,7 +726,19 @@ text(67, 0.03, "40%\n(0.40)", cex = 1, col = IMSCOL["black", "full"])
 
 In this case, the lower tail probability is known (0.40), which can be shaded on the diagram.
 We want to find the observation that corresponds to the known probability of 0.4.
-As a first step in this direction, we determine the Z score associated with the $40^{th}$ percentile.
+We can find the observation in two different ways: using the height curve seen above or using the Z score associated with the standard normal curve centered at zero with a standard deviation of one.
+
+If you have access to software (like R, code seen below) that allows you to specify the mean and standard deviation of the normal curve, you can calculate the observed value on the curve (i.e., Yousef's height) directly.
+
+```{r}
+qnorm(0.4, mean = 70, sd = 3.3)
+```
+
+Yousef is 69.2 inches tall.
+That is, Yousef is about 5'9'' (this is notation for 5-feet, 9-inches).
+
+Without access to flexible software, you will need the information given by a standard normal curve (a normal curve centered at zero with a standard deviation of one).
+First, determine the Z score associated with the $40^{th}$ percentile.
 
 Because the percentile is below 50%, we know $Z$ will be negative.
 Statistical software provides the $Z$ value to be $-0.25.$
@@ -738,17 +750,19 @@ qnorm(0.4, mean = 0, sd = 1)
 
 Knowing $Z_{Yousef}=-0.25$ and the population parameters $\mu=70$ and $\sigma=3.3$ inches, the Z score formula can be set up to determine Yousef's unknown height, labeled $x_{Yousef}$:
 
-$$ -0.25 = Z_{Yousef} = \frac{x_{Yousef} - \mu}{\sigma} = \frac{x_{Yousef} - 70}{3.3} $$
+$$ -0.253 = Z_{Yousef} = \frac{x_{Yousef} - \mu}{\sigma} = \frac{x_{Yousef} - 70}{3.3} $$
 
-Solving for $x_{Yousef}$ yields the height 69.18 inches.
-That is, Yousef is about 5'9'' (this is notation for 5-feet, 9-inches).
+Solving for $x_{Yousef}$ yields the height 69.2 inches.
+Again, Yousef is about 5'9''.
 :::
 
 ::: {.workedexample data-latex=""}
 What is the adult male height at the $82^{nd}$ percentile?
 
 ------------------------------------------------------------------------
 
+In order to practice using Z scores, we will use the standard normal curve to solve the problem.
+
 Again, we draw the figure first.
 
 ```{r}
@@ -764,7 +778,7 @@ And calculate the Z value associated with the $82^{nd}$ percentile:
 
 ```{r}
 #| echo: true
-qnorm(0.82, m = 0, s = 1)
+qnorm(0.82, mean = 0, sd = 1)
 ```
 
 Next, we want to find the Z score at the $82^{nd}$ percentile, which will be a positive value (because the percentile is bigger than 50%).
@@ -776,6 +790,8 @@ This yields 73.04 inches or about 6'1'' as the height at the $82^{nd}$ percentil
 :::
 
 ::: {.guidedpractice data-latex=""}
+Using Z scores, answer the following questions.
+
 (a) What is the $95^{th}$ percentile for SAT scores?\
 (b) What is the $97.5^{th}$ percentile of the male heights? As always with normal probability problems, first draw a picture.[^13-foundations-mathematical-11]
 :::
@@ -785,6 +801,8 @@ This yields 73.04 inches or about 6'1'' as the height at the $82^{nd}$ percentil
     (a) We look for 0.95 in the probability portion (middle part) of the normal probability table, which leads us to row 1.6 and (about) column 0.05, i.e., $Z_{95}=1.65.$ Knowing $Z_{95}=1.65,$ $\mu = 1500,$ and $\sigma = 300,$ we setup the Z score formula: $1.65 = \frac{x_{95} - 1500}{300}.$ We solve for $x_{95}$: $x_{95} = 1995.$ (b) Similarly, we find $Z_{97.5} = 1.96,$ again setup the Z score formula for the heights, and calculate $x_{97.5} = 76.5.$
 
 ::: {.guidedpractice data-latex=""}
+Using Z scores, answer the following questions.
+
 (a) What is the probability that a randomly selected male adult is at least 6'2'' (74 inches)?\
 (b) What is the probability that a male adult is shorter than 5'9'' (69 inches)?[^13-foundations-mathematical-12]
 :::
@@ -848,7 +866,7 @@ That is, the probability of being between 5'9'' and 6'2'' is 0.5048.
 :::
 
 ::: {.guidedpractice data-latex=""}
-What percent of SAT takers get between 1500 and 2000?[^13-foundations-mathematical-13]
+Find the percent of SAT takers who earn between 1500 and 2000.[^13-foundations-mathematical-13]
 :::
 
 [^13-foundations-mathematical-13]: This is an abbreviated solution.