Q8: An insurance company crashed four cars of the same model at 5 miles per hour. The costs of the repairs for each of the four crashes were $411, $436, $482 and $238. Compute the mean, medianand modecost of repair.

Q10: The scores and their percent of the final grade for a statisticsstudent are given. What is the student’s weighted mean score?

Q12: Find the range, mean, varianceand standard deviationof the sample dataset.

Q14: Heights of men on a baseball team have a bell-shaped distribution with a mean of 185 cm and a standard deviation of 7cm. Using the empirical rule, what is the approximate percentage of the men between the following values?

Q15: From a sample with n = 32, the mean duration of a geyser’s eruptions is 3.31 min and the standard deviation is 0.64 minutes. Using Chebychev’s Theorem, determine at least how many of the eruptions lasted between 2.03 and 4.559 min.

Q16: A student’s score on an actuarial exam is in the 78th percentile. What can you conclude about the student’s exam score?

Q17: Find the five-number summary, and draw a box-and-whisker plot of the data.

Q18: The data show the number of vacation days used in a recent year by a sample of 12 employees. Find the data set’s first, second and third quartiles. Draw a box-and-whisker plot that represents the data set.

Q19: The midpoints A, B, and C are marked on the histogram. Match them to the indicated scores. Which scores, if any, would be considered unusual?

Q20: The mean for statisticstest scores is 62 and the standard deviation is 6.0. or the biologytest scores the mean is 23 and the standard deviation is 3.8. The test score’s of a student who took both tests are given below.

See the attached file.

Q8:

There are four data: $411, $436, $482 and $238

Mean=(411+436+482+238)/4=391.75. so A is the right choice.

To figure out the median, we first organize the data in order: 238, 411, 436 and 482.

Median=(411+436)/2 (for even number, here there are 4 data, median is the average of middle two numbers)= 423.5. so A is right.

Note: for odd number of ordered data (suppose there are n data), median is the data in the (n+1)/2th position. Since the mode is the most frequent number, these dataset does not have the mode. So B is right.

Q10:

To find out the weighed mean, we multiple each data with its corresponding weighed ratio:

Weighed mean=82*0.2+83*0.15+96*0.15+99*0.25+87*0.25=89.75.

Q12:

12, 13, 17, 18, 6, 20, 11, 10, …

The solution discusses elementary statistics including mean, median and mode.