QUESTION IMAGE
Question
examples: percentages to raw data
- on the most recent test, a student scored in the 40th percentile. the mean of the test scores was an 85 and the standard deviation was 1.5. what was the student’s score?
- an automobile dealer finds that the average price of a previously owned vehicle is $8,256. he decides to sell cars that will appeal to the middle 60% of the market in terms of price. find the maximum and minimum prices of the cars the dealer will sell. the standard deviation is $1,150 and the variable is normally distributed.
mean (μ) = 8,256
std = $1,150
percent = 60%
min - (z, scores+
minimum =
maximum =
Step1: Find the z - score for the 40th percentile
Using a standard normal distribution table (z - table), the z - score corresponding to the 40th percentile (0.4000) is approximately $z=-0.25$.
Step2: Use the z - score formula
The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the raw score, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know $\mu = 85$, $\sigma=1.5$ and $z=-0.25$. Rearranging the formula for $x$ gives $x=\mu + z\sigma$.
Step3: Calculate the student's score
Substitute the values into the formula: $x = 85+(-0.25)\times1.5=85 - 0.375 = 84.625$.
for problem 8:
Step1: Determine the z - scores for the middle 60%
The middle 60% leaves 20% in the two tails (10% in each tail). The z - score corresponding to the 10th percentile (0.1000) is approximately $z_1=-1.28$ and the z - score corresponding to the 90th percentile (0.9000) is approximately $z_2 = 1.28$.
Step2: Use the z - score formula for the minimum price
The z - score formula is $z=\frac{x-\mu}{\sigma}$. Rearranging for $x$ gives $x=\mu+z\sigma$. For the minimum price, with $\mu = 8256$, $\sigma = 1150$ and $z=-1.28$, we have $x_1=8256+(-1.28)\times1150=8256-1472 = 6784$.
Step3: Use the z - score formula for the maximum price
Using the same formula $x=\mu + z\sigma$, with $z = 1.28$, $\mu = 8256$ and $\sigma=1150$, we get $x_2=8256 + 1.28\times1150=8256+1472=9728$.
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84.625