BE Computer Engineering (IOE, TU) Probability and Statistics (IOE, SH 602 / ENSH 301) Question Paper 2078 Nepal

Using mid-values $x$ and $f$ (total $N=100$):

(a) Mean, Median, Mode

Mean $=\dfrac{\sum fx}{N}=\dfrac{4860}{100}=48.6$ hrs.

Median: $N/2=50$. Cumulative frequencies: 12, 35, 70, 90, 100. The 50th item lies in class 40–60 (median class), $L=40$, $cf=35$, $f=35$, $h=20$.

$$\text{Median}=L+\frac{N/2-cf}{f}\,h=40+\frac{50-35}{35}\times20=40+8.57=48.57\text{ hrs}.$$

Mode: modal class is 40–60 (highest $f=35$), $L=40$, $f_1=35$, $f_0=23$, $f_2=20$, $h=20$.

$$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h=40+\frac{35-23}{70-23-20}\times20=40+\frac{12}{27}\times20=48.89\text{ hrs}.$$

(b) Standard deviation and CV

Using the step-deviation method ($h=20$):

$$\sigma=h\sqrt{\frac{\sum fd^2}{N}-\left(\frac{\sum fd}{N}\right)^2}=20\sqrt{\frac{131}{100}-\left(\frac{-7}{100}\right)^2}=20\sqrt{1.31-0.0049}=20\sqrt{1.3051}=22.85\text{ hrs}.$$ $$\text{CV}=\frac{\sigma}{\bar{x}}\times100=\frac{22.85}{48.6}\times100=47.0\%.$$

A CV of about 47% is fairly high, so the lifetimes show considerable variability — the components are not very consistent in lifetime.

(c) Karl Pearson coefficient of skewness

$$Sk=\frac{\text{Mean}-\text{Mode}}{\sigma}=\frac{48.6-48.89}{22.85}=-0.013.$$

The value is very close to 0 (slightly negative), so the distribution is almost symmetrical, with a very slight negative (left) skew.

Axioms of probability

For a sample space $S$ and any event $A$:

1. $P(A)\ge 0$ (non-negativity).
2. $P(S)=1$ (certainty).
3. If $A_1,A_2,\dots$ are mutually exclusive, $P(A_1\cup A_2\cup\cdots)=P(A_1)+P(A_2)+\cdots$ (additivity).

Theorem of total probability: If $B_1,B_2,\dots,B_n$ form a partition of $S$ (mutually exclusive and exhaustive) with $P(B_i)>0$, then for any event $A$:

$$P(A)=\sum_{i=1}^{n}P(B_i)\,P(A\mid B_i).$$

(a) Mutually exclusive vs independent

• Mutually exclusive: events cannot occur together, $P(A\cap B)=0$. Example: getting a head and a tail on a single coin toss.
• Independent: occurrence of one does not affect the other, $P(A\cap B)=P(A)\,P(B)$. Example: outcomes of two separate coin tosses.

Note: two events with non-zero probability cannot be both mutually exclusive and independent.

(b) Bayes' theorem for $M_2$

Let $D$ = item is defective. Given: $P(M_1)=0.25,\;P(M_2)=0.35,\;P(M_3)=0.40$ and $P(D\mid M_1)=0.05,\;P(D\mid M_2)=0.04,\;P(D\mid M_3)=0.02$.

$$P(D)=0.25(0.05)+0.35(0.04)+0.40(0.02)=0.0125+0.0140+0.0080=0.0345.$$ $$P(M_2\mid D)=\frac{P(M_2)P(D\mid M_2)}{P(D)}=\frac{0.0140}{0.0345}=0.4058.$$

So the probability the defective item came from $M_2$ is about 0.406 (40.6%).

(c) Probability of a non-defective item

$$P(\text{non-defective})=1-P(D)=1-0.0345=0.9655\;(96.55\%).$$

Continuous random variable & normal distribution

A continuous random variable takes any value in an interval; its distribution is described by a probability density function $f(x)\ge0$ with $\int_{-\infty}^{\infty}f(x)\,dx=1$ and $P(a\le X\le b)=\int_a^b f(x)\,dx$.

Properties of the normal distribution $N(\mu,\sigma^2)$:

• Bell-shaped and symmetric about $\mu$; mean = median = mode = $\mu$.
• Total area under the curve = 1; the curve is asymptotic to the $x$-axis.
• About 68%, 95% and 99.7% of values lie within $1\sigma,2\sigma,3\sigma$ of $\mu$.
• Defined by $f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}$; standardized by $Z=\frac{X-\mu}{\sigma}$.

(a) Normal approximation to binomial

The binomial $B(n,p)$ may be approximated by a normal distribution when:

• $n$ is large,
• $p$ is not too close to 0 or 1 (so the distribution is roughly symmetric), and
• in practice both $np\ge5$ and $n(1-p)\ge5$. Then $X\approx N(np,\,np(1-p))$ (with a continuity correction).

(b) Workers with wages between Rs. 69 and Rs. 72

$\mu=70,\;\sigma=5$. $Z_1=\frac{69-70}{5}=-0.20$, $Z_2=\frac{72-70}{5}=0.40$.

$$P(69<X<72)=P(-0.20<Z<0.40)=P(0<Z<0.20)+P(0<Z<0.40)=0.0793+0.1554=0.2347.$$

Number of workers $=1000\times0.2347\approx\textbf{235 workers}.$

(c) Lowest wage of the highest-paid 100 workers

Highest-paid 100 of 1000 = top 10%, so $P(X>x)=0.10$, i.e. area between mean and $x$ is $0.40$. From the table $P(0<Z<1.28)=0.3997\approx0.40$, so $Z=1.28$.

$$x=\mu+Z\sigma=70+1.28\times5=70+6.4=\textbf{Rs. }76.4.$$

The highest-paid 100 workers earn at least about Rs. 76.40 per week.

Computations ($n=8$)

$\bar{X}=48/8=6$, $\bar{Y}=398/8=49.75$.

(a) Karl Pearson correlation coefficient

$$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^2-(\sum X)^2][n\sum Y^2-(\sum Y)^2]}}.$$

Numerator $=8(2575)-48(398)=20600-19104=1496$. $n\sum X^2-(\sum X)^2=8(316)-48^2=2528-2304=224$. $n\sum Y^2-(\sum Y)^2=8(21062)-398^2=168496-158404=10092$.

$$r=\frac{1496}{\sqrt{224\times10092}}=\frac{1496}{\sqrt{2260608}}=\frac{1496}{1503.5}=0.995.$$

There is a very strong positive linear correlation between study hours and marks.

(b) Regression equations

$$b_{YX}=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}=\frac{1496}{224}=6.679.$$ $$b_{XY}=\frac{n\sum XY-\sum X\sum Y}{n\sum Y^2-(\sum Y)^2}=\frac{1496}{10092}=0.1483.$$

Y on X: $Y-\bar{Y}=b_{YX}(X-\bar{X})\Rightarrow Y-49.75=6.679(X-6)$

$$\boxed{Y=6.679X+9.674.}$$

X on Y: $X-\bar{X}=b_{XY}(Y-\bar{Y})\Rightarrow X-6=0.1483(Y-49.75)$

$$\boxed{X=0.1483Y-1.378.}$$

(c) Estimate for 10 hours & link to $r$

For $X=10$: $Y=6.679(10)+9.674=\textbf{76.5 marks (approx.)}$.

Relation: $r=\pm\sqrt{b_{YX}\cdot b_{XY}}=\sqrt{6.679\times0.1483}=\sqrt{0.9905}=0.995$, which matches part (a) (positive sign since both regression coefficients are positive).

(a) Value of $k$

Since $\sum P(X)=1$:

$$k+2k+3k+2k+k=9k=1\Rightarrow k=\frac{1}{9}.$$

So the distribution is $P(0)=\tfrac19,P(1)=\tfrac29,P(2)=\tfrac39,P(3)=\tfrac29,P(4)=\tfrac19$.

(b) Mean and variance

$$E(X)=\sum xP(x)=\frac{0+2+6+6+4}{9}=\frac{18}{9}=2.$$ $$E(X^2)=\sum x^2P(x)=\frac{0+2+12+18+16}{9}=\frac{48}{9}=5.333.$$ $$\text{Var}(X)=E(X^2)-[E(X)]^2=5.333-4=\textbf{1.333}\;\left(=\tfrac{12}{9}\right).$$

Thus $E(X)=2$ and $\text{Var}(X)=\tfrac43\approx1.33$ (the distribution is symmetric about $X=2$).

Binomial with $n=8$, $p=0.1$, $q=0.9$. $P(X=r)=\binom{8}{r}(0.1)^r(0.9)^{8-r}$.

(a) Exactly 2 defective

$$P(X=2)=\binom{8}{2}(0.1)^2(0.9)^6=28\times0.01\times0.531441=\textbf{0.1488}.$$

(b) At most 1 defective; mean & variance

$$P(X=0)=(0.9)^8=0.4305,\qquad P(X=1)=\binom{8}{1}(0.1)(0.9)^7=8\times0.1\times0.4783=0.3826.$$ $$P(X\le1)=0.4305+0.3826=\textbf{0.8131}.$$

Mean $=np=8\times0.1=0.8$. Variance $=npq=8\times0.1\times0.9=0.72$.

Poisson with mean $\lambda=3$ calls/min, $P(X=r)=\dfrac{e^{-3}3^r}{r!}$, $e^{-3}=0.0498$.

(a) No call in a minute

$$P(X=0)=e^{-3}=\textbf{0.0498}.$$

(b) More than 2 calls

$$P(X>2)=1-[P(0)+P(1)+P(2)].$$

$P(1)=e^{-3}\cdot3=0.1494$, $\;P(2)=e^{-3}\cdot\dfrac{9}{2}=0.0498\times4.5=0.2241$.

$$P(X\le2)=0.0498+0.1494+0.2241=0.4233.$$ $$P(X>2)=1-0.4233=\textbf{0.5767}.$$

Given $n=64$, $\bar{x}=52$, $s=8$.

(a) Point vs interval estimate

• A point estimate is a single value used to estimate a population parameter (e.g. $\bar{x}=52$ for $\mu$).
• An interval estimate gives a range of values, with a stated confidence level, within which the parameter is expected to lie (e.g. a 95% confidence interval). It conveys the precision/uncertainty of the estimate, which a point estimate does not.

(b) 95% confidence interval for $\mu$

Standard error $=\dfrac{s}{\sqrt{n}}=\dfrac{8}{\sqrt{64}}=1$. With $z_{0.025}=1.96$:

$$\text{CI}=\bar{x}\pm z\cdot\frac{s}{\sqrt n}=52\pm1.96(1)=52\pm1.96.$$ $$\boxed{(50.04,\;53.96)}.$$

Interpretation: we are 95% confident that the true population mean lies between 50.04 and 53.96 units; i.e. if many such samples were taken, about 95% of the resulting intervals would contain $\mu$.

Given $\mu_0=1800$ N, $n=50$, $\bar{x}=1770$, $s=100$.

(a) Hypotheses

• $H_0:\mu\ge1800$ N (claim is true) vs $H_1:\mu<1800$ N.
• This is a one-tailed (left-tailed) test.

(b) Test at 5% level

Test statistic (large sample $z$-test):

$$z=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}=\frac{1770-1800}{100/\sqrt{50}}=\frac{-30}{14.142}=-2.121.$$

Critical value (left-tailed, 5%): $-z_{0.05}=-1.645$.

Since $z=-2.121<-1.645$, the test statistic falls in the rejection region, so we reject $H_0$.

Conclusion: at the 5% level there is sufficient evidence that the mean breaking strength is less than 1800 N; the manufacturer's claim is not justified.

Hypotheses: $H_0$: the die is fair (all faces equally likely) vs $H_1$: the die is not fair.

Under $H_0$, expected frequency for each face $=120/6=20$.

$$\chi^2=\sum\frac{(O-E)^2}{E}=2.30.$$

Degrees of freedom $=k-1=6-1=5$. Critical value $\chi^2_{0.05,5}=11.07$.

Since calculated $\chi^2=2.30<11.07$, we do not reject $H_0$.

Conclusion: there is no significant evidence against fairness — the die may be regarded as fair at the 5% level.

Given $P(A)=0.5$, $P(B)=0.4$, $P(A\cup B)=0.7$.

(a) $P(A\cap B)$ and $P(A\mid B)$

$$P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.5+0.4-0.7=\textbf{0.2}.$$ $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.2}{0.4}=\textbf{0.5}.$$

(b) Independence and $P(\bar A\cap\bar B)$

$P(A)\,P(B)=0.5\times0.4=0.2=P(A\cap B)$. Since $P(A\cap B)=P(A)P(B)$, the events $A$ and $B$ are independent. (Also $P(A\mid B)=0.5=P(A)$ confirms this.)

By De Morgan's law:

$$P(\bar A\cap\bar B)=P(\overline{A\cup B})=1-P(A\cup B)=1-0.7=\textbf{0.3}.$$

(a) First four central moments

The $r$-th central moment about the mean is $\mu_r=E[(X-\bar X)^r]$:

• $\mu_1=E(X-\bar X)=0$ (always zero).
• $\mu_2=E[(X-\bar X)^2]=$ variance.
• $\mu_3=E[(X-\bar X)^3]$ — measures skewness.
• $\mu_4=E[(X-\bar X)^4]$ — measures kurtosis.

(b) Conversion from moments about 5

Given $\mu_1'=2,\;\mu_2'=20,\;\mu_3'=40,\;\mu_4'=250$.

$$\mu_2=\mu_2'-(\mu_1')^2=20-4=\textbf{16}.$$ $$\mu_3=\mu_3'-3\mu_2'\mu_1'+2(\mu_1')^3=40-3(20)(2)+2(8)=40-120+16=\textbf{-64}.$$ $$\mu_4=\mu_4'-4\mu_3'\mu_1'+6\mu_2'(\mu_1')^2-3(\mu_1')^4=250-4(40)(2)+6(20)(4)-3(16)$$ $$=250-320+480-48=\textbf{362}.$$

Skewness: $\beta_1=\dfrac{\mu_3^2}{\mu_2^3}=\dfrac{(-64)^2}{16^3}=\dfrac{4096}{4096}=1$, and $\gamma_1=\dfrac{\mu_3}{\mu_2^{3/2}}=\dfrac{-64}{64}=-1$.

Since $\mu_3<0$ (and $\gamma_1=-1$), the distribution is negatively (left) skewed.