Question

1. according to the truecar.com website, the asking prices for cars of a certain model are approximately normally distributed with mean $35,987 and standard deviation $607.50.

a. the factory invoice price for this model is $35,625. about what percent of cars have asking prices below the factory invoice price?
b. the manufacturers suggested retail price (msrp) for this car model is at the 98th percentile of the distribution of asking price. find the value of the msrp.

Explanation:

Part (a)

Step 1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the dataset, $\mu$ is the mean, and $\sigma$ is the standard deviation.
Given $\mu = 35987$, $\sigma=607.50$ (note: there was a typo in the original hand - written work, the standard deviation is $607.50$ not $5007.50$), and $x = 35625$.
$z=\frac{35625 - 35987}{607.50}=\frac{- 362}{607.50}\approx - 0.596$

Step 2: Find the area to the left of the z - score

Using the standard normal distribution table (z - table), the area to the left of $z=- 0.60$ (a close approximation) is approximately 0.2236 or 22.36%. If we use a more precise calculation for $z=-0.596$, we can use the formula for the standard normal distribution or a calculator. The cumulative distribution function for the standard normal distribution $\Phi(z)$ gives the probability that a standard normal variable is less than $z$. Using a calculator or more precise z - table, $\Phi(- 0.596)\approx0.2766$ (wait, I made a mistake in the sign earlier. Wait, $x = 35625$, $\mu=35987$, so $x-\mu=35625 - 35987=-362$, $z=\frac{-362}{607.50}\approx - 0.596$. The area to the left of $z = - 0.596$: looking up in the z - table, for $z=-0.60$, the area is 0.2743, for $z = - 0.59$, the area is 0.2776. Using linear interpolation between $z=-0.59$ and $z=-0.60$:
The difference in z - scores: $-0.59-(-0.60) = 0.01$
The difference in areas: $0.2776 - 0.2743=0.0033$
Our z - score is $z=-0.596$, which is $0.006$ above $z=-0.60$ (since $-0.596-(-0.60)=0.004$? Wait, no: $z=-0.596$ is $0.004$ less than $z=-0.59$ (because $-0.59-(-0.596) = 0.006$). Wait, let's recast:
Let $z=-0.59 + t$, where $t=-0.006$. The area at $z=-0.59$ is $A_1 = 0.2776$, the area at $z=-0.60$ is $A_2 = 0.2743$. The slope of the area with respect to z is $\frac{A_2 - A_1}{-0.60-(-0.59)}=\frac{0.2743 - 0.2776}{-0.01}=\frac{- 0.0033}{-0.01}=0.33$ per unit z.
So for $z=-0.596$ (which is $t=-0.006$ from $z=-0.59$), the area is $A=A_1+t\times0.33=0.2776-0.00198 = 0.27562$. So approximately 27.56% (I had a sign error earlier, the z - score is negative, but the area to the left of a negative z - score is the probability that the variable is less than that value. So my initial mistake was in the standard deviation value. The correct standard deviation is $607.50$, not $5007.50$ as in the hand - written work. So with the correct $\sigma = 607.50$, the z - score calculation is different. Let's recalculate:

$\mu=35987$, $\sigma = 607.50$, $x = 35625$

$z=\frac{35625 - 35987}{607.50}=\frac{-362}{607.50}\approx - 0.596$

Using a standard normal calculator (like the one on a TI - 84 or an online calculator), the probability that $Z<-0.596$ is approximately $P(Z < - 0.596)=\Phi(-0.596)\approx0.2766$ or 27.66%.

Part (b)

Step 1: Find the z - score corresponding to the 98th percentile

The 98th percentile means that 98% of the data is less than or equal to this value. So we need to find $z$ such that $P(Z < z)=0.98$.
Using the standard normal distribution table or a calculator, the z - score corresponding to a cumulative probability of 0.98 is approximately $z = 2.05$ (more precisely, using a calculator, $z=\Phi^{-1}(0.98)\approx2.0537$)

Step 2: Use the z - score formula to find $x$ (MSRP)

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $z$ is the z - score, $x$ is the value we want to find, $\mu$ is the mean, and $\sigma$ is the standard deviation.
We know $z\approx2.0537$, $\mu = 35987$, $\sigma=607.50$.
Rearranging the formula for $x$: $x=\mu+z\times\sigma$
$x = 35987+2.0537\times607.50$
First, calculate $2.0537\times607.50\approx2.0537\times600 + 2.0537\times7.50=1232.22+15.40275 = 1247.62275$
Then, $x=35987 + 1247.62275=37234.62275\approx\$37234.62$ (if we use $z = 2.05$, $2.05\times607.50 = 1245.375$, $x=35987+1245.375 = 37232.375\approx\$37232.38$)

Answer:

Approximately 22.66% (or using more precise z - table values, around 22.36%)