The following data give the no of passengers travelling by Boeing 747 from one city to another in one week.
320, 290, 265, 300, 270, 315
Calculate the mean and standard deviation
Solution
Mean μ=∑XiN\mu = \dfrac{\sum X_i}{N}μ=N∑Xi
Standard deviation σ=∑(Xi−μ)2N\sigma=\sqrt{\dfrac{\sum (X_i-\mu)^2}{N}}σ=N∑(Xi−μ)2
(a) Mean
μ=265+270+290+300+315+3206\mu =\frac{265+270+290+300+315+320}{6}μ=6265+270+290+300+315+320
μ=17606=293.33\mu =\dfrac{1760}{6}=293.33μ=61760=293.33
(b) Standard deviation
XX−μ(X−μ)2265−28.33802.77270−23.33544.44290−3.3311.113006.6744.4431521.67469.4432026.67711.11∑2583.33\def\arraystretch{1.5}\begin{array}{c:c:c}X&X-\mu&(X-\mu)^2\\\hline265&-28.33&802.77\\\hdashline270&-23.33&544.44\\\hdashline290&-3.33&11.11\\\hdashline300&6.67&44.44\\\hdashline315&21.67&469.44\\\hdashline320&26.67&711.11\\\hline\sum &&2583.33\\\hline\end{array}X265270290300315320∑X−μ−28.33−23.33−3.336.6721.6726.67(X−μ)2802.77544.4411.1144.44469.44711.112583.33
σ=2583.336\sigma=\sqrt{\dfrac{2583.33}{6}}σ=62583.33
σ=20.75\sigma=20.75σ=20.75
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