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Sagot :
Mean: $22,000
Standard deviation: $2,000
a)
For $18,000 we have:
[tex]Z_{18000}=\frac{18000-22000}{2000}=-2[/tex]For $26,000, we have:
[tex]Z_{26000}=\frac{26000-22000}{2}=2[/tex]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:
[tex]\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}[/tex]Therefore, the middle 99.7% ofcar costs are between $16,000 and $28000
c)
[tex]Z_{24000}=\frac{24000-22000}{2}=1[/tex]Therefore, we have:
[tex]P(Z<1)=50\%+\frac{68\%}{2}=84\%[/tex]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:
[tex]P(Z>2)=50\%+\frac{95\%}{2}=97.5\%[/tex]The probability a car will cost more than $26,000 is 97.5%
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