Question

Consider the data regarding car costs. The mean is $22,000 and the standard deviation is $2,000.a) Not everyone pays the same price for the same model of car. Use the 68-95-99.7% Rule to find what percentage ofbuyers paid between $18,000 and $26,000.b) The middle 99.7% of car costs are between what values?c) What is the probability a car will cost less than $24,000?d) What is the probability a car will cost more than $26,000?

Answer 1

Mean: $22,000

Standard deviation: $2,000

a)

For $18,000 we have:

$Z_{18000}=\frac{18000-22000}{2000}=-2$

For $26,000, we have:

$Z_{26000}=\frac{26000-22000}{2}=2$

Therefore, according to the 68-95-99,7% rule, the percentage of buyers that paid between $18,000 and $26,000 is 95%.

b)

To find the values between which lies the middle 99.7% of car costs, we have:

$\begin{gathered} Z_-=-3=\frac{x_1-22000}{2000} \\ x_1=22000-6000=\text{ \ }16000 \\ Z_+=3=\frac{x_2-22,000}{2,000} \\ x_2=22000+6000=\text{ \ }28000 \end{gathered}$

Therefore, the middle 99.7% ofcar costs are between $16,000 and $28000

c)

$Z_{24000}=\frac{24000-22000}{2}=1$

Therefore, we have:

$P(Z<1)=50\%+\frac{68\%}{2}=84\%$

The probability a car will cost less than $24,000 is 84%

d)

We already know that, for $26,000, Z = 2. Therefore, we have:

$P(Z>2)=50\%+\frac{95\%}{2}=97.5\%$

The probability a car will cost more than $26,000 is 97.5%