Mathematical Statistics: Difficult Problems with Solutions

Solution:
Express [tex]x[/tex] from the transformation:
[tex]y=\frac{4x-21}{10}\iff x=\frac{10y+21}{4}=2.5y+5.25[/tex]
so every original value is [tex]x_i=2.5y_i+5.25[/tex]. For a linear transformation [tex]x_i=ay_i+b[/tex] we have [tex]\bar{x}=a\bar{y}+b[/tex], hence
[tex]x_i-\bar{x}=a\left(y_i-\bar{y}\right)\ \Rightarrow\ \left(x_i-\bar{x}\right)^2=a^2\left(y_i-\bar{y}\right)^2[/tex]
Summing over all [tex]i[/tex] and dividing by the same divisor gives [tex]S_x^2=a^2S_y^2[/tex]: the shift [tex]b[/tex] does not change the spread, and the factor [tex]a[/tex] enters squared. Therefore
[tex]S_x^2=2.5^2\times 1.1=6.25\times 1.1=6.875[/tex]
(The mean [tex]\bar{x}=2.5\times 3.1+5.25=13[/tex] is not needed for the variance.)

Solution:
Let [tex]X[/tex] be the TOP score and [tex]Y[/tex] the score on the new module; the new score is [tex]X+Y[/tex]. Denote the variance by [tex]D[/tex]. For any two random variables
[tex]D(X+Y)=D(X)+D(Y)+2\operatorname{cov}(X,Y)[/tex]
By the definition of the correlation coefficient, [tex]\operatorname{cov}(X,Y)=r\,\sigma_X\sigma_Y[/tex], where [tex]\sigma_X=\sqrt{D(X)}[/tex] and [tex]\sigma_Y=\sqrt{D(Y)}[/tex] are the standard deviations. Here [tex]r=\frac{1}{3}[/tex], so
[tex]\operatorname{cov}(X,Y)=\frac{1}{3}\times \sqrt{18}\times \sqrt{2}=\frac{1}{3}\times 6=2[/tex]
Therefore
[tex]D(X+Y)=18+2+2\times 2=24[/tex]
The means [tex]56[/tex] and [tex]4[/tex] are not needed. The standard deviation grows from [tex]\sqrt{18}\approx 4.24[/tex] to [tex]\sqrt{24}\approx 4.90[/tex], so the spread does increase.

Solution:
The sum of the seven known values is [tex]111[/tex], so the mean is [tex]\frac{111+x}{8}[/tex]. In ascending order the known values are [tex]6, \ 8, \ 14, \ 15, \ 18, \ 20, \ 30[/tex]. Together with [tex]x[/tex] there are [tex]8[/tex] values, so the median is the average of the fourth and the fifth value in ascending order, and these depend on the position of [tex]x[/tex].
Case [tex]x\le 14[/tex]: the fourth and fifth values are [tex]14[/tex] and [tex]15[/tex], the median is [tex]14.5[/tex], and
[tex]\frac{111+x}{8}=14.5\iff 111+x=116\iff x=5[/tex]
This value satisfies [tex]x\le 14[/tex]. Case [tex]14\le x\le 18[/tex]: the two middle values are [tex]15[/tex] and [tex]x[/tex], the median is [tex]\frac{x+15}{2}[/tex], and
[tex]\frac{111+x}{8}=\frac{x+15}{2}\iff 111+x=4x+60\iff x=17[/tex]
This value lies in [tex]\left[14, \,18\right][/tex]. Case [tex]x\ge 18[/tex]: the two middle values are [tex]15[/tex] and [tex]18[/tex], the median is [tex]16.5[/tex], and
[tex]\frac{111+x}{8}=16.5\iff 111+x=132\iff x=21[/tex]
Indeed, [tex]21\ge 18[/tex]. Thus [tex]x\in\{5, \,17, \,21\}[/tex], and the greatest possible value is [tex]21[/tex]. Check: for [tex]x=21[/tex] the median is [tex]\frac{15+18}{2}=16.5[/tex] and the mean is [tex]\frac{132}{8}=16.5[/tex].

Solution:
Let the results of the first experiment in ascending order be [tex]x_1\le x_2\le\ldots\le x_{100}[/tex]. The number of data is even, so the median is the average of the terms with indices [tex]50[/tex] and [tex]51[/tex]:
[tex]\sum_{i=1}^{100}x_i=100\times 55=5500,\qquad \frac{x_{50}+x_{51}}{2}=49[/tex]
The results of the second experiment in ascending order are [tex]x_1-1, \ \ldots, \ x_{50}-1, \ x_{51}+2, \ \ldots, \ x_{100}+2[/tex] (the order is preserved, because [tex]x_{50}-1\lt x_{50}+2\le x_{51}+2[/tex]). Their sum and their mean are
[tex]5500-50\times 1+50\times 2=5550,\qquad \frac{5550}{100}=55.5[/tex]
Their median is the average of the terms with indices [tex]50[/tex] and [tex]51[/tex]:
[tex]\frac{(x_{50}-1)+(x_{51}+2)}{2}=\frac{x_{50}+x_{51}}{2}+\frac{1}{2}=49+0.5=49.5[/tex]
Hence the mean exceeds the median by
[tex]55.5-49.5=6[/tex]

Solution:
The sample size is [tex]n=40[/tex], so the upper quartile corresponds to the position [tex]\frac{3n}{4}=30[/tex]. The cumulative frequencies of the intervals are
[tex]4, \ 9, \ 15, \ 19, \ 24, \ 29, \ 34, \ 40[/tex]
The first cumulative frequency that reaches [tex]30[/tex] is [tex]34[/tex], so [tex]Q_3[/tex] lies in the interval [tex][3.8, \,4.0)[/tex]. For this interval the lower bound is [tex]L=3.8[/tex], the width is [tex]h=0.2[/tex] (the widths are not all equal, so the width of this particular interval must be used), the frequency is [tex]m=5[/tex] and the cumulative frequency before it is [tex]F=29[/tex]. By linear interpolation
[tex]Q_3=L+\frac{\frac{3n}{4}-F}{m}\times h=3.8+\frac{30-29}{5}\times 0.2=3.84[/tex]

Solution:
Let [tex]x[/tex] be the number of boys and [tex]y[/tex] the number of girls. The means are
[tex]\bar{x}=\frac{150}{6}=25,\qquad \bar{y}=\frac{120}{6}=20[/tex]
The deviations from the means are
[tex]x_i-\bar{x}:\ -5, \ 0, \ 0, \ 5, \ -5, \ 5[/tex]
[tex]y_i-\bar{y}:\ -5, \ 0, \ 0, \ 0, \ -5, \ 10[/tex]
Hence
[tex]S_{xy}=\sum(x_i-\bar{x})(y_i-\bar{y})=25+0+0+0+25+50=100[/tex]
[tex]S_{xx}=\sum(x_i-\bar{x})^2=25+0+0+25+25+25=100[/tex]
[tex]S_{yy}=\sum(y_i-\bar{y})^2=25+0+0+0+25+100=150[/tex]
The factor [tex]\frac{1}{n}[/tex] in the covariance and in both variances cancels, so
[tex]r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{100}{\sqrt{100\times 150}}=\frac{100}{50\sqrt{6}}=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}\approx 0.816[/tex]
The correlation is strong and positive: in months with more boys born, more girls tend to be born as well.

Solution:
The observations are independent, so the likelihood function is the product of the densities. Since [tex]\sum x_i=2+3+5+4+7=21[/tex],
[tex]L(\lambda)=\prod_{i=1}^{5}\lambda e^{-\lambda x_i}=\lambda^5e^{-\lambda\sum x_i}=\lambda^5e^{-21\lambda}[/tex]
Taking logarithms and setting the derivative equal to zero:
[tex]\ln L(\lambda)=5\ln\lambda-21\lambda,\qquad \frac{d}{d\lambda}\ln L(\lambda)=\frac{5}{\lambda}-21=0\iff \lambda=\frac{5}{21}[/tex]
This is a maximum, because [tex]\frac{d^2}{d\lambda^2}\ln L(\lambda)=-\frac{5}{\lambda^2}\lt 0[/tex]. Therefore
[tex]\hat{\lambda}=\frac{n}{\sum x_i}=\frac{1}{\bar{x}}=\frac{5}{21}\approx 0.238[/tex]
i.e. the estimate is the reciprocal of the sample mean lifetime [tex]\bar{x}=4.2[/tex] years.

Solution:
The sample proportion is [tex]\hat{p}=\frac{16}{25}=0.64[/tex]. Under [tex]H_0[/tex] the number of infected sailors is binomial with [tex]n=25[/tex] and [tex]p_0=0.5[/tex]. Since [tex]np_0=n(1-p_0)=12.5\ge 5[/tex], the normal approximation may be used, and the standard error is computed from the value [tex]p_0[/tex] assumed by [tex]H_0[/tex]:
[tex]\sqrt{\frac{p_0(1-p_0)}{n}}=\sqrt{\frac{0.25}{25}}=0.1[/tex]
Therefore
[tex]z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}=\frac{0.64-0.5}{0.1}=1.4[/tex]
The test is right-sided and [tex]z=1.4\lt 1.645[/tex], so [tex]H_0[/tex] is not rejected at level [tex]0.05[/tex]: the data do not give sufficient grounds to claim that the probability of infection is greater than [tex]0.5[/tex].

Solution:
Here [tex]n=10[/tex]. The sample mean is
[tex]\bar{x}=\frac{23+26+15+14+10+15+23+21+23+30}{10}=\frac{200}{10}=20[/tex]
The deviations from the mean are [tex]3, \ 6, \ -5, \ -6, \ -10, \ -5, \ 3, \ 1, \ 3, \ 10[/tex], and the sum of their squares is
[tex]\sum_{i=1}^{10}(x_i-\bar{x})^2=9+36+25+36+100+25+9+1+9+100=350[/tex]
so the unbiased sample variance and the sample standard deviation are
[tex]s^2=\frac{350}{9}\approx 38.89, \qquad s\approx 6.236[/tex]
The population is normal and its variance is unknown, so the statistic [tex]\frac{\bar{X}-\mu}{s/\sqrt{n}}[/tex] has Student's [tex]t[/tex]-distribution with [tex]n-1=9[/tex] degrees of freedom, and the [tex]95\%[/tex] confidence interval is [tex]\bar{x}\pm t_{0.975, \,9}\frac{s}{\sqrt{n}}[/tex]. We have
[tex]\frac{s}{\sqrt{n}}=\sqrt{\frac{350}{90}}=\frac{\sqrt{35}}{3}\approx 1.972[/tex]
and the margin of error is
[tex]t_{0.975, \,9}\times \frac{s}{\sqrt{n}}\approx 2.262\times 1.972\approx 4.46[/tex]
Hence the confidence interval (in thousands of dollars) is
[tex]\left(20-4.46 , 20+4.46\right)=\left(15.54 , 24.46\right)[/tex]