2.1-2.3 Practice Problems

The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.

What is the probability that a person waits fewer than 12.5 minutes?

(write your answer to 4 decimal places)

Tags

CCSS.HSS.MD.B.6

CCSS.HSS.MD.B.7

The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.

On average, how long must a person wait? (Find the mean)

(leave out the units in your answer)

The waiting time for the train that leaves every 30 minutes is uniformly distributed from 0 to 30 minutes. Find the probability that a person arriving at a random time will wait between 10 and 15 minutes.

• 0.5

• 0.667

• 0.167

• 0.333

Tags

CCSS.7.SP.C.7A

For more than a century, doctors have been telling patients that a normal body temperature is 98.6°F (37.0 degrees Celsius). This value dates back to a study done by Carl Wunderlich in the mid-1800s. More recently, researchers conducted a study to determine whether the "accepted" value for normal body temperature is accurate. They collected body temperatures (in degrees Celsius) from 130 healthy individuals. A dotplot and numerical summaries of the data are given. At what percentile is 36.0 degrees Celsius?

• 3.08th Percentile

• 3.85th Percentile

• 27.7th Percentile

• 32.5th Percentile

Tags

CCSS.6.SP.B.5C

Find the 10th percentile of the distribution of body temperatures.

• 36.5

• 36.2

• 36.3

• 36.0

Tags

CCSS.6.SP.B.5C

Michael's temperature was 37.5 degrees Celsius. Calculate Michael's standardized score (z-score)

• 52.93

• 1.71

• -52.93

• -1.71

Tags

CCSS.HSS.ID.A.4

Which of the following is the correct interpretation of the z-score calculated in the last problem. The correct z-score was 1.71.

• Michael's temperature is higher than 1.71% of the other temperatures

• Michael's temperature is 1.71 standard deviations above the mean

• Michael's temperature is 1.71 degrees above the mean

• Michael's temperature is 1.71% above the mean

• Mean = 64.249

  SD = 0.826

• Mean = 64.249

  SD = 0.407

• Mean = 66.249

  SD = 0.407

• Mean = 66.249

  SD = 0.7326

The next step to convert to Fahrenheit would be to add 32 to all of the values. Knowing that the mean from the previous question was 66.249 and the standard deviation was 0.7326, what would the new mean and standard deviation be after adding 32 to all values?

• Mean = 98.249

  SD = 0.7326

• Mean = 98.249

  SD = 32.7326

• Mean = 66.249

  SD = 32.7326

• Mean = 66.249

  SD = 0.7326

The scores on a certain college exam have a mean μ = 80 and standard deviation σ = 4. Duane scores an 84 on this exam.

The scores on another college exam have a mean μ = 85 and standard deviation σ = 8. Debbie scores an 90 on this exam.

Relative to their own exam score distributions, who scored higher on their exam?

• Debbie because her exam score is higher

• Debbie since the standard deviation for her exam is greater

• Duane because his score is more standard deviations above the mean

• Debbie because her score is more standard deviations above the mean

Tags

CCSS.HSS.ID.A.4