Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 3 - Statistics for Describing, Exploring, and Comparing Data - Review - Review Exercises: 1

Answer

a)1559.6 b)1550 c)No mode. d)1569.5 e)145 f)53.38 g)2849.42. h)1517.5. i)1606.5.

Work Step by Step

a) The mean can be counted by summing all the data and dividing it by the number of data: $\frac{1550+1642+1538+1497+1571}{5}=1559.6.$ b)The median is the average of the 1 or 2 (here 1) middle data: (1550)/1=1550. c) The mode is the data occuring the most times but here all data occurs one time, therefore there is no mode. d) The midrange is the average of the highest and the lowest data: (1497+1642)/2=1569.5. e) The range is the difference of the highest and the lowest data: 1642-1497=145. f)Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(1550-1559.6)^2+...+(1571-1559.6)^2}{4}}=53.38.$ g) variance=$(standard \ deviation)^2=53.38^2=2849.42.$ h)$Q_{1}$ is the value under which 25% of the values lie. Hence $Q_{1}=\frac{1497+1538}{2}=1517.5.$ i)$Q_{3}$ is the value above which 25% of the values lie. Hence $Q_{3}=\frac{1571+1642}{2}=1606.5.$
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