Question

Exercise 4. Density Curves a. In each row of this table, assume the distribution is normal. Knowing any two of the mean, standard deviation, Q1 or Q3 is enough to determine the other two. Complete the table by filling in numbers for 0-(V). (You may use pnorm and qnorm in Rand also use R as a calculator.) Mean Q. Q SD 5 10 (0) (ii) (1) (iv) 20 80 b. A study conducted by the Kaiser Foundation compared blood pressures for women having different numbers of children. Below are sketches of the density curves for blood pressure (measured in mm) in women with 2 or 4 children. I. Is the number of children a woman has a numerical or categorical variable in the graphic Support your answer. Write answer here ii. The caption for the y-axis is missing. What do the numbers on the y axis tell us? (Hint look at the brief introduction to density histograms at the end of the slidedeck titled Continuous Distributions) Write answer here lll. True or false and explain the increase in the number of children has an additive effect on blood pressure. Write true/false first and then support your answer iv. Roughly, what is the height of the density curve for women with 4 children? Show your work. Write answer first and then support with reason v. Blood pressures above 140 mmHg are considered unusually high. Which group has a higher probability of exceeding 140- the mothers of 2 children or 4 children? Explain how you know this. (You do not need to do calculations - just use your eyeballs) Write which group and then support with reason

Answer 1

a)

Let X follows Normal distribution with mean $\mu$ and standard deviation $\sigma$ .

$Z=\frac{X-\mu}{\sigma}$ follows standard normal distribution.

1st Quartile Q1 is such that, P[ X<Q1] = 0.25

and 3rd Quartile Q3 is such that, P[ X<Q3 ] = 0.75

Let $\mu$ = 5 and $\sigma$ = 10

P[ X<Q1] = 0.25

= PI - P < 1 - H = PIZ < = 0.25 0 0 0

Let us first find the Q1 and Q2 for standard normal distribution.

#qnorm(0.25,0,1)

q1 = -0.6744

#qnorm(0.75,0,1)

q3 = 0.6744

01-1 -0.6744 4 01 = 5 - 0.6744 * 10 = -1.744 0

$\frac{Q3-\mu}{\sigma}=0.6744\Rightarrow Q3=5+0.6744*10 = 11.744$

Or you can directly find Q1 and Q3 using following R commands

#qnorm(0.25,5,10)

#qnorm(0.75,5,10)

Now given, Q1=20 and Q3=80

From above part of the solution we can have that,

$\frac{Q1-\mu}{\sigma}=-0.6744\Rightarrow \mu-0.6744\sigma=20$ _ _ _ (1)

$\frac{Q3-\mu}{\sigma}=0.6744\Rightarrow \mu+0.6744\sigma=80$ _ _ _ (2)

Subtract equation (1) from (2)

$1.3488\sigma = 60\Rightarrow \sigma=44.484$

Substitute $\sigma$ in equation (1)

$\mu=20+0.6744\sigma=30$

Hence, $\mu=30$ and $\sigma=44.484$

To answer the questions in part (b) you haven't provided the necessary graph, that is the density curves. Hence, we cannot answer the questions.

I hope you find the solution. Feel free to ask doubt, if any in the comment section.