Question

Problem 1 The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 32 and a standard deviation of about 12 Suppose that 18 individuals are randomly chosen. Let X = average percent of fat calories consumed for a sample of size 18 Enter numbers as integers or fractions in "p/q" form, or as decimals accurate to nearest 0.01 a.(20) X ~ B ( 32 , 12v18 ) . b. (20) Use the mean and standard deviation of X to determine the z value for X 35 c. (20) mustrate PX < 35) as an area by adjusting the slider along the horizontal axis to control the z value. 0.00 0.300 0.2 0.1

Answer 1

Let X be the percent of fat calories that a person in America consumes each day. X is normally distributed with mean $\mu=32$ and standard deviation $\sigma=12$

A sample of size n=18 individuals is randomly chosen. Let $\bar{X}$ be the average percent of fat calories consumed for a sample of size n=18.

We know that $\bar{X}$ is normally distributed with mean $\mu_{\bar{x}}=\mu=32$ and standard deviation (or also called standard error)  

$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{12}{\sqrt{18}}=2.83$

a) ans: $\bar{X}\sim \mathcal{N}(32,12/\sqrt{18})$ fractions in"p/q" format.

or in decimals as

$\bar{X}\sim \mathcal{N}(32,2.83)$

b) The z value for $\bar{X}=35$ is

$z=\frac{35-\mu_{\bar{x}}}{\sigma_{\bar{x}}}=\frac{35-32}{2.83}=1.06$

c) The value of z is 1.06. We need to move the slider along to the right till we get to z=1.06 as shown below. The shaded region to the left of the vertical line would be the region corresponding to $\bar{X}<35$ or Z<1.06

0.00 0.300 に1.06 0.2 0.1 OFill left Fll right

d) The probability that the average consumption for 18 people in America is less than 35 is

$\begin{align*} P(\bar{X}<35)=P(Z<1.06)\quad\text{We represent 35 using its z value}\\ \end{align*}$

Using the standard normal table we read the value for z=1.06 as 0.3554

The value of P(Z<1.06) = 0.5+0.3553=0.8553

ans: $\begin{align*} P(\bar{X}<35)=0.86 \end{align*}$

e) Let $\begin{align*} Q_1 \end{align*}$ be the first quartile for the average percent of fat calories.

We know that 25% of the observations are less than the first quartile $\begin{align*} Q_1 \end{align*}$ . That is the probability that the average consumption for 18 people in America is less than $\begin{align*} Q_1 \end{align*}$ is 0.25

$\begin{align*} P(\bar{X}<Q_1)=0.25 \end{align*}$

First we will find the z value for which P(Z<z) = 0.25. We know that the probability that Z is less than 0 is 0.5. That is P(Z<0) = 0.5

Hence z has to be less than zero (negative) for the probability to be less than 0.5.

Hence we will write the required probability as

$\begin{align*} &P(Z<-z)=0.25\\ \implies &P(Z>z)=0.25\quad\text{this is due to the symmetry of normal distribution}\\ \implies &1-P(Z<z)=0.25\\ \implies &P(Z<z)=1-0.25\\ \implies &P(Z<z)=0.75\\ \end{align*}$

Now we will use the standard normal table and find that for z=0.67 the area under the normal curve is (0.5+0.2486)=0.7486. Since this is close enough to 0.75, we will use this value. (For more accurate value, for z=0.68 the area under the normal curve is 0.5+0.2517=0.7517. Now we interpolate and get a more accurate 0.674)

That is P(Z<0.67) = 0.75 or P(Z<-0.67) = 0.25

Now we will get the value of $\begin{align*} Q_1 \end{align*}$ by equating the z value of $\begin{align*} Q_1 \end{align*}$ to -0.67

$\begin{align*} \frac{Q_1-\mu_{\bar{x}}}{\sigma_{\bar{x}}}=\frac{Q_1-32}{2.83} \end{align*}$ is the z value, which we will equate to -0.67

$\begin{align*}&\frac{Q_1-32}{2.83}=-0.67\\ \implies &Q_1=32-2.83\times 0.67\\ \implies &Q_1=30.10 \end{align*}$

(If we use a more accurate value of 0.674 we will get the value of Q1=30.09)