a man recorded the speeds of cars driving past his house, where the speed limit was 25 mph. the mean of 100…
a man recorded the speeds of cars driving past his house, where the speed limit was 25 mph. the mean of 100 readings was 28.75 mph with a standard deviation of 3.69 mph. (a) how many standard deviations from the mean would a car going the speed limit be? (b) which would be more unusual, a car traveling 40 mph or one going 17 mph? (round to two decimal places as needed.) (a) a car traveling at the speed limit is 1.02 standard deviations from the mean. (b) choose the correct answer below and fill in the answer box(es) to complete your choice. (round to two decimal places as needed.) a. the car traveling 17 mph is more unusual. it is standard deviations from the mean, while the car traveling 40 mph is standard deviations from the mean. b. the car traveling 40 mph is more unusual. it is standard deviations from the mean, while the car traveling 17 mph is standard deviations from the mean. c. both cars are equally unusual. both cars are standard deviations from the mean.
Answer
Explanation:
Step1: Calculate z - scores for 40 mph and 17 mph
The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 28.75$ mph and $\sigma=3.69$ mph. For $x = 40$ mph: $z_1=\frac{40 - 28.75}{3.69}=\frac{11.25}{3.69}\approx3.05$. For $x = 17$ mph: $z_2=\frac{17 - 28.75}{3.69}=\frac{- 11.75}{3.69}\approx - 3.18$.
Step2: Analyze the unusualness
In general, a z - score with $|z|>2$ is considered unusual. Since $|z_1|\approx3.05>2$ and $|z_2|\approx3.18>2$, both cars are equally unusual.
Answer:
a) For 40 mph: 3.05; For 17 mph: - 3.18 b) C. Both cars are equally unusual. Both cars are standard deviations from the mean.