12ஆம் வகுப்பு கணிதவியல் — நிகழ்தகவு பரவல்கள்: Book Back MCQs with Answers & Explanations

The mean is \(E(X)=\displaystyle\int_{1}^{\infty} x\cdot\dfrac{2}{x^{3}}\,dx=\int_{1}^{\infty}\dfrac{2}{x^{2}}\,dx=\left[-\dfrac{2}{x}\right]_{1}^{\infty}=2\), which is finite. For the variance we first need \(E(X^{2})=\displaystyle\int_{1}^{\infty} x^{2}\cdot\dfrac{2}{x^{3}}\,dx=\int_{1}^{\infty}\dfrac{2}{x}\,dx=\big[\,2\ln x\,\big]_{1}^{\infty}\), and this integral diverges. Since \(E(X^{2})\) is infinite, \(\operatorname{Var}(X)=E(X^{2})-\big(E(X)\big)^{2}\) cannot be evaluated. Hence the mean exists but the variance does not.

The shorter piece is uniformly distributed on \((0,l)\). Its mean is \(E(X)=\displaystyle\int_{0}^{l} x\cdot\dfrac{1}{l}\,dx=\dfrac{1}{l}\cdot\dfrac{l^{2}}{2}=\dfrac{l}{2}\). Next, \(E(X^{2})=\displaystyle\int_{0}^{l} x^{2}\cdot\dfrac{1}{l}\,dx=\dfrac{1}{l}\cdot\dfrac{l^{3}}{3}=\dfrac{l^{2}}{3}\). Therefore \(\operatorname{Var}(X)=E(X^{2})-\big(E(X)\big)^{2}=\dfrac{l^{2}}{3}-\dfrac{l^{2}}{4}=\dfrac{l^{2}}{12}\). So the mean is \(\dfrac{l}{2}\) and the variance is \(\dfrac{l^{2}}{12}\).

Each face of a fair die has probability \(\dfrac{1}{6}\). The player gains ₹36 only when a 6 appears, and loses ₹\(k^{2}\) for the faces \(k=1,2,3,4,5\). Hence the expected winning is \(E=\dfrac{1}{6}(36)+\dfrac{1}{6}\displaystyle\sum_{k=1}^{5}(-k^{2})=6-\dfrac{1}{6}(1+4+9+16+25)=6-\dfrac{55}{6}=\dfrac{36-55}{6}=-\dfrac{19}{6}\). The negative value shows the game is unfavourable to the player.

We count ordered pairs \((a,b)\) with \(a\in\{1,2,3,4,5,6\}\), \(b\in\{1,2,3,4\}\) and \(a+b=7\). These are \((3,4),(4,3),(5,2)\) and \((6,1)\). A pair such as \((7,0)\) is impossible because the four-sided die has no face \(0\), and taking \(a\lt 3\) would require \(b\gt 4\). So exactly \(4\) outcomes give a sum of \(7\).

For a binomial distribution the standard deviation is \(\sigma=\sqrt{npq}\) with \(q=1-p\). Here \(q=1-0.8=0.2\), so \(\sigma=\sqrt{25\times 0.8\times 0.2}=\sqrt{4}=2\).

Suppose the coin shows \(i\) heads in \(n\) tosses; then it shows \(n-i\) tails. The difference is \(X=i-(n-i)=2i-n\). As \(i\) runs through \(0,1,2,\dots,n\), the attainable values of \(X\) are exactly \(2i-n\).

For a uniform density the total area must equal \(1\), so \(\displaystyle\int_{a}^{b}\dfrac{1}{12}\,dx=\dfrac{b-a}{12}=1\), which forces \(b-a=12\). Checking each pair: \(12-0=12\) (valid), \(17-5=12\) (valid), \(19-7=12\) (valid), but \(24-16=8\neq 12\) (invalid). Hence \(a=16,\ b=24\) cannot occur.

Choosing a student at random makes a larger bus more likely to be picked, so \(X\) is size-biased: \(E(X)=\displaystyle\sum(\text{bus size})\times\dfrac{\text{bus size}}{160}=\dfrac{42^{2}+36^{2}+34^{2}+48^{2}}{160}=\dfrac{1764+1296+1156+2304}{160}=\dfrac{6520}{160}=40.75\). Choosing a driver at random makes every bus equally likely, so \(E(Y)=\dfrac{42+36+34+48}{4}=\dfrac{160}{4}=40\). Hence \(E(X)=40.75\) and \(E(Y)=40\).

Because the draws are made with replacement, each draw independently gives an ace with probability \(\dfrac{4}{52}=\dfrac{1}{13}\). Write \(X=X_{1}+X_{2}\), where \(X_{i}=1\) if the \(i\)-th card is an ace and \(0\) otherwise; each \(X_{i}\) is a Bernoulli variable with mean \(\dfrac{1}{13}\). By the linearity of expectation, \(E(X)=E(X_{1})+E(X_{2})=\dfrac{1}{13}+\dfrac{1}{13}=\dfrac{2}{13}\).

Each question is answered correctly with probability \(p=\dfrac{1}{3}\), so \(X\sim B\!\left(5,\dfrac{1}{3}\right)\) with \(q=\dfrac{2}{3}\). Then \(P(X\ge 4)=P(X=4)+P(X=5)=\dbinom{5}{4}\!\left(\dfrac{1}{3}\right)^{4}\!\left(\dfrac{2}{3}\right)+\left(\dfrac{1}{3}\right)^{5}=\dfrac{10}{243}+\dfrac{1}{243}=\dfrac{11}{243}\).

For \(B(n,p)\) we have \(E(X)=np\) and \(\operatorname{Var}(X)=npq\). The condition \(E(X)=3\operatorname{Var}(X)\) gives \(np=3npq\), so \(1=3q\), i.e. \(q=\dfrac{1}{3}\) and \(p=\dfrac{2}{3}\). With \(n=4\), \(P(X=0)=\dbinom{4}{0}p^{0}q^{4}=\left(\dfrac{1}{3}\right)^{4}=\dfrac{1}{81}\).

For \(B(n,p)\), \(np=6\) and \(npq=2.4\). Dividing, \(q=\dfrac{npq}{np}=\dfrac{2.4}{6}=0.4=\dfrac{2}{5}\), so \(p=\dfrac{3}{5}\) and \(n=\dfrac{6}{p}=10\). Therefore \(P(X=5)=\dbinom{10}{5}\left(\dfrac{3}{5}\right)^{5}\left(\dfrac{2}{5}\right)^{5}\).

A cumulative distribution function must satisfy \(F(\infty)=1\) and, for continuity at the origin, \(F(0)=0\). As \(x\to\infty\), \(e^{-x}\to 0\), so \(F(\infty)=a=1\). At \(x=0\), \(F(0)=a+b=0\), giving \(b=-1\). Thus \(F(x)=1-e^{-x}\) for \(x\ge 0\), the standard exponential distribution, so \(a=1\) and \(b=-1\).

For \(B(n,p)\), the mean is \(np=2k\) and the variance is \(npq=k\). Dividing, \(q=\dfrac{npq}{np}=\dfrac{k}{2k}=\dfrac{1}{2}\), so \(p=\dfrac{1}{2}\). With \(n=8\), \(np=8\times\dfrac{1}{2}=4\); since \(np=2k\), we get \(2k=4\), i.e. \(k=2\).

A continuous random variable takes any value in an interval (it is measured), whereas a discrete one takes isolated, countable values (it is counted). The lifetime of a bulb (I) and the duration of a phone call (III) are measured quantities that can take any value in an interval, so they are continuous. The number of defective items (II) is a count, hence discrete. Therefore I and III are the continuous random variables.

The probabilities of a pmf must add up to \(1\). Here \(\displaystyle\sum_{x=1}^{\infty} a\left(\dfrac{1}{2}\right)^{x}=a\sum_{x=1}^{\infty}\left(\dfrac{1}{2}\right)^{x}\). The geometric series \(\displaystyle\sum_{x=1}^{\infty}\left(\dfrac{1}{2}\right)^{x}=\dfrac{\tfrac{1}{2}}{1-\tfrac{1}{2}}=1\). Hence \(a\cdot 1=1\), giving \(a=1\).

First find \(k\) from \(\sum f(x)=1\): \(k+2k+3k+4k+5k=15k=1\), so \(k=\dfrac{1}{15}\). Then \(E(X)=\displaystyle\sum x\,f(x)=(-2)k+(-1)(2k)+0(3k)+(1)(4k)+(2)(5k)=(-2-2+0+4+10)k=10k=\dfrac{10}{15}=\dfrac{2}{3}\).

For a Bernoulli variable, \(p=\) mean \(=0.4\), so \(q=1-p=0.6\) and \(\operatorname{Var}(X)=pq=0.4\times 0.6=0.24\). Using \(\operatorname{Var}(aX+b)=a^{2}\operatorname{Var}(X)\) with \(a=2\) (the constant \(-3\) does not affect the variance): \(\operatorname{Var}(2X-3)=2^{2}\times 0.24=4\times 0.24=0.96\).

With \(n=6\), \(P(X=4)=\dbinom{6}{4}p^{4}q^{2}\) and \(P(X=2)=\dbinom{6}{2}p^{2}q^{4}\). Since \(\dbinom{6}{4}=\dbinom{6}{2}=15\), the relation \(9P(X=4)=P(X=2)\) becomes \(9p^{2}=q^{2}\). Taking positive roots, \(3p=q=1-p\), so \(4p=1\) and \(p=\dfrac{1}{4}=0.25\).

Selling to a customer is a “success” with probability \(p=\dfrac{1}{20}\), so \(q=\dfrac{19}{20}\), and there are \(n=3\) customers, giving \(X\sim B\!\left(3,\dfrac{1}{20}\right)\). Then \(P(X=2)=\dbinom{3}{2}\left(\dfrac{1}{20}\right)^{2}\left(\dfrac{19}{20}\right)=3\cdot\dfrac{1}{400}\cdot\dfrac{19}{20}=\dfrac{57}{8000}\).