ap stats: normal distribution calculations - practice\nmr. wilcox spends a full - day at gingerman raceway…

ap stats: normal distribution calculations - practice\nmr. wilcox spends a full - day at gingerman raceway testing some new tires on his honda civic. for 400 laps, he records his lap time in minutes. at the end of the day, he plots the distribution of times and realizes that it follows an approximately normal distribution with a mean of 1.84 and a standard deviation of 0.07.\n1. label the values 1, 2, and 3 standard deviations above and below the mean.\n2. what percent of the lap times are less than 1.70 minutes?\n3. what percent of the lap times are greater than 1.77 minutes?\n4. what percent of the lap times are between 1.70 and 1.91 minutes?\n5. what lap time would be in the slowest 2.5% of all lap times?

ap stats: normal distribution calculations - practice\nmr. wilcox spends a full - day at gingerman raceway testing some new tires on his honda civic. for 400 laps, he records his lap time in minutes. at the end of the day, he plots the distribution of times and realizes that it follows an approximately normal distribution with a mean of 1.84 and a standard deviation of 0.07.\n1. label the values 1, 2, and 3 standard deviations above and below the mean.\n2. what percent of the lap times are less than 1.70 minutes?\n3. what percent of the lap times are greater than 1.77 minutes?\n4. what percent of the lap times are between 1.70 and 1.91 minutes?\n5. what lap time would be in the slowest 2.5% of all lap times?

Answer

Explanation:

Step1: Calculate values for standard - deviation intervals

Let $\mu = 1.84$ be the mean and $\sigma=0.07$ be the standard deviation. 1 standard deviation above the mean: $\mu+\sigma=1.84 + 0.07=1.91$ 1 standard deviation below the mean: $\mu-\sigma=1.84-0.07 = 1.77$ 2 standard deviations above the mean: $\mu + 2\sigma=1.84+2\times0.07=1.98$ 2 standard deviations below the mean: $\mu-2\sigma=1.84 - 2\times0.07=1.70$ 3 standard deviations above the mean: $\mu+3\sigma=1.84+3\times0.07 = 2.05$ 3 standard deviations below the mean: $\mu-3\sigma=1.84-3\times0.07=1.63$

Step2: Calculate z - scores and use the standard normal table for question 2

The z - score formula is $z=\frac{x-\mu}{\sigma}$. For $x = 1.70$, $z=\frac{1.70 - 1.84}{0.07}=\frac{- 0.14}{0.07}=-2$. Looking up the z - score in the standard normal table, the area to the left of $z = - 2$ is $0.0228$ or $2.28%$.

Step3: Calculate z - scores and use the standard normal table for question 3

For $x = 1.77$, $z=\frac{1.77-1.84}{0.07}=\frac{-0.07}{0.07}=-1$. The area to the left of $z=-1$ is $0.1587$. The area to the right (lap times greater than 1.77) is $1 - 0.1587=0.8413$ or $84.13%$.

Step4: Calculate z - scores and use the standard normal table for question 4

For $x = 1.70$, $z_1=\frac{1.70 - 1.84}{0.07}=-2$. For $x = 1.91$, $z_2=\frac{1.91-1.84}{0.07}=1$. The area to the left of $z_1=-2$ is $0.0228$ and the area to the left of $z_2 = 1$ is $0.8413$. The area between them is $0.8413-0.0228 = 0.8185$ or $81.85%$.

Step5: Calculate the lap time for the slowest 2.5% for question 5

The z - score corresponding to the area of $0.025$ to the left is $z=-1.96$. Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we can solve for $x$. Rearranging gives $x=\mu+z\sigma$. Substituting $\mu = 1.84$, $z=-1.96$ and $\sigma = 0.07$, we get $x=1.84+(-1.96)\times0.07=1.84 - 0.1372=1.7028$ minutes.

Answer:

    • 1 standard deviation above the mean: 1.91
    • 1 standard deviation below the mean: 1.77
    • 2 standard deviations above the mean: 1.98
    • 2 standard deviations below the mean: 1.70
    • 3 standard deviations above the mean: 2.05
    • 3 standard deviations below the mean: 1.63
  2. $2.28%$
  3. $84.13%$
  4. $81.85%$
  5. 1.7028 minutes