Mathematical Statistics: Problems with Solutions

Solution:
The total number of grades is
[tex]n=2+4+6+10+1=23[/tex]
Since [tex]n[/tex] is odd, the median is the middle term of the ordered series, namely the term with number
[tex]\frac{n+1}{2}=\frac{23+1}{2}=12[/tex]
Order the grades increasingly and accumulate the frequencies: the grade [tex]2[/tex] occupies places [tex]1[/tex] and [tex]2[/tex], the grade [tex]3[/tex] places [tex]3[/tex] to [tex]6[/tex], the grade [tex]4[/tex] places [tex]7[/tex] to [tex]12[/tex], the grade [tex]5[/tex] places [tex]13[/tex] to [tex]22[/tex] and the grade [tex]6[/tex] place [tex]23[/tex]. Hence the [tex]12[/tex]th term is [tex]4[/tex]:
[tex]Me=4[/tex]

Solution:
The sum of [tex]m[/tex] numbers equals [tex]m[/tex] times their mean. Hence the sums of the ten, of the nine and of the eight numbers are
[tex]10\times 17=170,\qquad 9\times 16=144,\qquad 8\times 15=120[/tex]
The first removed number is the difference of the first two sums, and the second removed number is the difference of the last two sums:
[tex]x=170-144=26,\qquad y=144-120=24[/tex]
Their product is
[tex]xy=26\times 24=624[/tex]

Solution:
Split the sum into two sums:
[tex]\sum_{i=1}^{n}(x_i-\bar{x})=\sum_{i=1}^{n}x_i-\sum_{i=1}^{n}\bar{x}[/tex]
The number [tex]\bar{x}[/tex] does not depend on [tex]i[/tex], so the second sum consists of [tex]n[/tex] equal terms; and by the definition of the mean the first sum also equals [tex]n\bar{x}[/tex]:
[tex]\sum_{i=1}^{n}\bar{x}=n\bar{x},\qquad \bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i\ \Rightarrow\ \sum_{i=1}^{n}x_i=n\bar{x}[/tex]
Therefore
[tex]\sum_{i=1}^{n}(x_i-\bar{x})=n\bar{x}-n\bar{x}=0[/tex]
which is what we had to prove. Thus the positive and the negative deviations from the mean always cancel each other exactly; this is why measures of spread use squared (or absolute) deviations.

Solution:
The estimate of the expected value is the sample mean:
[tex]\bar{x}=\frac{2.1+3.2+4.6+2.9+3.5+4.1}{6}=\frac{20.4}{6}=3.4[/tex]
The deviations from the mean are [tex]-1.3, \ -0.2, \ 1.2, \ -0.5, \ 0.1, \ 0.7[/tex], and the sum of their squares is
[tex]\sum_{i=1}^{6}(x_i-\bar{x})^2=1.69+0.04+1.44+0.25+0.01+0.49=3.92[/tex]
The unbiased estimate of the variance divides by [tex]n-1=5[/tex], not by [tex]n[/tex] (dividing by [tex]n[/tex] gives a biased estimate, because one degree of freedom is used up by estimating the mean):
[tex]s^2=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2=\frac{3.92}{5}=0.784[/tex]

Solution:
The standard deviation is
[tex]\sigma=\sqrt{4}=2[/tex]
By Chebyshev's inequality, which holds for every data set whatever its distribution, for every [tex]k\gt 1[/tex] at least a fraction [tex]1-\frac{1}{k^2}[/tex] of the data lie in the interval [tex]\left[\bar{x}-k\sigma, \bar{x}+k\sigma\right][/tex]. This guaranteed fraction is at least [tex]\frac{8}{9}[/tex] exactly when
[tex]1-\frac{1}{k^2}\ge\frac{8}{9}\iff k^2\ge 9\iff k\ge 3[/tex]
so the narrowest such interval is obtained for [tex]k=3[/tex]:
[tex]\left[60-3\times 2, 60+3\times 2\right]=\left[54, 66\right][/tex]
Note that [tex]k[/tex] multiplies the standard deviation [tex]2[/tex], not the variance [tex]4[/tex].

Solution:
The point estimate of the energy is the sample mean:
[tex]\bar{x}=\frac{260+216+259+206+265+284+291+229+249}{9}=\frac{2259}{9}=251[/tex]
The population is normal and [tex]\sigma[/tex] is known, so [tex]\frac{\bar{X}-\mu}{\sigma}\sqrt{n}[/tex] has the standard normal distribution, and the [tex]95\%[/tex] confidence interval is [tex]\bar{x}\pm z_{0.975}\frac{\sigma}{\sqrt{n}}[/tex]. The margin of error is
[tex]z_{0.975}\times \frac{\sigma}{\sqrt{n}}=1.96\times \frac{10}{3}\approx 6.53[/tex]
Therefore
[tex]\mu\in\left[251-6.53, 251+6.53\right]=\left[244.47, 257.53\right][/tex]