0. Check the model The mean of the 100 car speeds in Exercise 26 was 23.84 mph,

Question

0. Check the model The mean of the 100 car speeds in Exercise 26 was 23.84 mph, with a standard deviation of 3.56 mph a) Using a Normal model, what values should border the middle 95% of all car speeds? b) Here are some summary statistics Percentile 100% 97.5% 90.0% 75.0% 50.0% 25.0% 10.0% 25% 0.0% Speed 34.060 30.976 28.978 25.785 Median 23.525 21.547 19.163 16.638 16.270 Max 03 Q1 Min From your answer in part a, how well does the model do in predicting those percentiles? Are you surprised? Explain.

Explanation / Answer

Answer to the question is as follows:

Mean of 100 cars is 23.94 mph and stdev is 3.56 mph

a. Values in middle 95% is

Xbar +/- Z*Sigma/sqrt(n)

= 23.94 +/- 1.96*3.56/sqrt(100) = 23.2422 to 24.6377

b.

The Max is 100 percentile i.e. 3 deviation above mean is 99.7%
Hence, 23.94 + 3*3.56 = 34.62 mph which is near 34.060

2 deviations more than mean is 97.5%
Hence, 23.94 + 2*3.56 = 31.06 mph which is near 30.976

Also, Median is close to the mean. Median of 23.525 is almost equal 23.84 mph

Also, Q1 and Q3 is 25 and 75th percentile value of 21.547 and 28.978 is which is close to 23.84 +/- 0.675*3.56 which is 21.437 and 26.243

Hence, all points fit the values given in the tables and therefore model is good in predicting these percenitles. I am not surprised. This because these values may have come from a model with approzximately the same population distribution.

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