BE Computer Engineering (Pokhara University) Probability and Statistics (PU, MTH 216) Question Paper 2078 Nepal

We use mid-points $x$ of each class; $N=\sum f = 40$.

(a) Mean, Median, Mode

Mean: $\bar x = \dfrac{\sum fx}{N} = \dfrac{1530}{40} = 38.25$ ms.

Median: $N/2 = 20$ lies in class 30–40 (c.f. just exceeds 20). With $L=30,\ F=11,\ f_m=12,\ h=10$:

$$\text{Median} = L + \frac{N/2 - F}{f_m}\,h = 30 + \frac{20-11}{12}\times 10 = 37.5 \text{ ms}.$$

Mode: modal class is 30–40 (highest $f=12$). With $f_1=12,\ f_0=7,\ f_2=9$:

$$\text{Mode} = L + \frac{f_1-f_0}{2f_1-f_0-f_2}\,h = 30 + \frac{12-7}{24-7-9}\times 10 = 30 + 6.25 = 36.25 \text{ ms}.$$

(b) Standard deviation and coefficient of variation

$$\sigma = \sqrt{\frac{\sum f(x-\bar x)^2}{N}} = \sqrt{\frac{7477.5}{40}} = \sqrt{186.94} = 13.67 \text{ ms}.$$ $$\text{C.V.} = \frac{\sigma}{\bar x}\times 100 = \frac{13.67}{38.25}\times 100 = 35.7\%.$$

Comment: A C.V. of about $35.7\%$ is fairly high, so the server's response times show considerable relative variability — the response is not very consistent; latency fluctuates substantially across requests.

(c) Karl Pearson's coefficient of skewness

$$S_k = \frac{\bar x - \text{Mode}}{\sigma} = \frac{38.25 - 36.25}{13.67} = +0.146.$$

Since $S_k > 0$ (and $\text{Mean} > \text{Median} > \text{Mode}$), the distribution is slightly positively (right) skewed: most response times are moderate, with a thin tail of longer response times pulling the mean above the mode.

Let $A,B,C$ be the events that a chip comes from suppliers A, B, C, and $D$ the event that the chip is defective.

$$P(A)=0.50,\ P(B)=0.30,\ P(C)=0.20,\quad P(D|A)=0.02,\ P(D|B)=0.03,\ P(D|C)=0.04.$$

(a) Theorems and prior/posterior probability

Theorem of total probability: If $A_1,\dots,A_n$ are mutually exclusive and exhaustive events with $P(A_i)>0$, then for any event $D$:

$$P(D)=\sum_{i=1}^{n} P(A_i)\,P(D|A_i).$$

Bayes' theorem: For any $A_k$,

$$P(A_k|D)=\frac{P(A_k)\,P(D|A_k)}{\sum_{i} P(A_i)\,P(D|A_i)}.$$

• The prior probability $P(A_k)$ is the probability assigned to a hypothesis before observing the evidence (here, the proportion of stock from each supplier).
• The posterior probability $P(A_k|D)$ is the revised probability of the hypothesis after the evidence $D$ (a defective chip) has been observed.

(b) Probability the chip is defective

$$P(D)=P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)$$ $$=0.50(0.02)+0.30(0.03)+0.20(0.04)=0.010+0.009+0.008=0.027.$$

So $P(D)=0.027$ (i.e. $2.7\%$).

(c) Probability it came from supplier C, given defective

$$P(C|D)=\frac{P(C)P(D|C)}{P(D)}=\frac{0.008}{0.027}=0.296.$$

(d) Most likely source of a defective chip

$$P(A|D)=\frac{0.010}{0.027}=0.370,\quad P(B|D)=\frac{0.009}{0.027}=0.333,\quad P(C|D)=0.296.$$

The largest posterior is $P(A|D)=0.370$, so a defective chip is most likely to have come from supplier A. Although A has the lowest defect rate, it supplies half of all chips, so it contributes the most defective units in absolute terms.

$n=8$. Computing sums:

$$\sum X=57,\ \sum Y=520,\ \sum XY=4130,\ \sum X^2=491,\ \sum Y^2=35938.$$ $$\bar X=57/8=7.125,\quad \bar Y=520/8=65.$$

(a) Karl Pearson coefficient of correlation

$$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^2-(\sum X)^2][n\sum Y^2-(\sum Y)^2]}}$$ $$=\frac{8(4130)-57(520)}{\sqrt{[8(491)-57^2][8(35938)-520^2]}}=\frac{33040-29640}{\sqrt{(3928-3249)(287504-270400)}}$$ $$=\frac{3400}{\sqrt{679\times 17104}}=\frac{3400}{3407.9}=0.998.$$

Interpretation: $r\approx +0.998$ indicates a very strong positive linear correlation — students who practise more hours score higher, almost perfectly linearly.

(b) Least-squares regression line of Y on X

$$b_{YX}=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}=\frac{3400}{679}=5.007.$$ $$a=\bar Y-b_{YX}\bar X=65-5.007(7.125)=65-35.68=29.32.$$

Regression line:

$$\boxed{\hat Y = 29.32 + 5.007\,X.}$$

(c) Prediction and coefficient of determination

For $X=10$ hours:

$$\hat Y = 29.32 + 5.007(10) = 29.32 + 50.07 = 79.4.$$

The expected score is about 79 out of 100.

Coefficient of determination: $r^2 = (0.998)^2 = 0.995$. This means about 99.5% of the variation in assessment scores is explained by the variation in practice hours through the fitted line; only $0.5\%$ is due to other factors. The very high $r^2$ confirms the line fits the data extremely well.

Given: $\mu_0=8000$ h (claim: at least 8000), $\bar x=7820$ h, $s=480$ h, $n=36$.

(a) Errors, level of significance, p-value

• Type I error ($\alpha$): rejecting the null hypothesis $H_0$ when it is actually true (a 'false alarm').
• Type II error ($\beta$): failing to reject $H_0$ when it is actually false (a 'missed detection').
• Level of significance $\alpha$: the maximum probability of committing a Type I error that we are willing to tolerate (here $0.05$); it fixes the rejection region.
• p-value: the probability, assuming $H_0$ is true, of obtaining a test statistic at least as extreme as the one observed. If p-value $< \alpha$, we reject $H_0$.

(b) Hypothesis test (5% level)

$$H_0:\mu \ge 8000 \quad\text{(claim true)}\qquad H_1:\mu < 8000 \quad\text{(left-tailed)}.$$

Since $n=36$ is large, use the $z$-test (using $s$ for $\sigma$):

$$z=\frac{\bar x-\mu_0}{s/\sqrt n}=\frac{7820-8000}{480/\sqrt{36}}=\frac{-180}{80}=-2.25.$$

Critical value for a left-tailed test at $\alpha=0.05$ is $-z_{0.05}=-1.645$.

Since $z=-2.25 < -1.645$, the test statistic falls in the rejection region (equivalently p-value $=P(Z<-2.25)\approx 0.0122 < 0.05$).

Conclusion: Reject $H_0$. There is sufficient evidence at the 5% level that the true mean lifetime is less than 8000 hours, so the manufacturer's claim is not justified.

(c) 95% confidence interval

$$\bar x \pm z_{0.025}\,\frac{s}{\sqrt n}=7820 \pm 1.96\times 80 = 7820 \pm 156.8.$$ $$\text{CI} = (7663.2,\ 7976.8) \text{ hours}.$$

The entire interval lies below 8000 hours; 8000 is not contained in it. This is consistent with part (b) — at the 5% level the data are incompatible with a mean of 8000 h or more, confirming the claim is not supported.

Drawing without replacement, so the second draw's probability is conditional on the first; we use the multiplication rule $P(E_1\cap E_2)=P(E_1)P(E_2|E_1)$, and for (c) the complement/addition idea.

(a) Both aces. There are 4 aces.

$$P(\text{both aces})=\frac{4}{52}\times\frac{3}{51}=\frac{12}{2652}=\frac{1}{221}\approx 0.00452.$$

(b) First king, then queen.

$$P=\frac{4}{52}\times\frac{4}{51}=\frac{16}{2652}=\frac{4}{663}\approx 0.00603.$$

(c) At least one spade. Easiest via the complement (no spade in either draw). There are 39 non-spades.

$$P(\text{no spade})=\frac{39}{52}\times\frac{38}{51}=\frac{1482}{2652}=\frac{19}{34}\approx 0.559.$$ $$P(\text{at least one spade})=1-\frac{19}{34}=\frac{15}{34}\approx 0.441.$$

Each bit errs independently with $p=0.01$, $n=100$. The number of errors $X$ is $\text{Binomial}(100,0.01)$. Since $n$ is large and $p$ is small with $np=1$ moderate, we approximate by the Poisson distribution with mean

$$\lambda = np = 100\times 0.01 = 1.$$

The approximation is appropriate because $n\ge 20$ (large), $p\le 0.05$ (small/rare event), and $np<10$, the standard conditions for the Poisson approximation to the binomial.

Poisson pmf: $P(X=k)=\dfrac{e^{-\lambda}\lambda^{k}}{k!}$ with $\lambda=1$.

(a) No error ($k=0$):

$$P(X=0)=e^{-1}=0.3679.$$

(b) At most two errors ($k=0,1,2$):

$$P(X\le 2)=e^{-1}\!\left(1+1+\tfrac{1}{2}\right)=e^{-1}(2.5)=2.5\times 0.3679=0.9197.$$

So about $36.8\%$ of packets are error-free and about $92.0\%$ contain at most two errors.

Let $X$ = bolt length, $X\sim N(\mu=50,\ \sigma=1.5)$. Standardise with $Z=\dfrac{X-\mu}{\sigma}$.

(a) Percentage acceptable (between 47.5 and 52.5 mm)

$$Z_1=\frac{47.5-50}{1.5}=-1.67,\qquad Z_2=\frac{52.5-50}{1.5}=+1.67.$$ $$P(47.5<X<52.5)=P(-1.67<Z<1.67)=2\,\Phi(1.67)-1.$$

From the standard normal table $\Phi(1.67)=0.9525$, so

$$P=2(0.9525)-1=0.9050.$$

About $90.5\%$ of bolts are acceptable.

(b) Expected number rejected out of 2000

Fraction rejected $=1-0.9050=0.0950$.

$$\text{Rejected}=2000\times 0.0950 \approx 190 \text{ bolts.}$$

So roughly 190 bolts per shift are expected to be rejected.

$f(x)=kx(2-x)$ for $0\le x\le 2$, else 0.

(a) Value of k

A pdf integrates to 1:

$$\int_0^2 kx(2-x)\,dx = k\int_0^2 (2x-x^2)\,dx = k\Big[x^2-\tfrac{x^3}{3}\Big]_0^2 = k\Big(4-\tfrac{8}{3}\Big)=k\cdot\tfrac{4}{3}=1.$$ $$\Rightarrow k=\frac{3}{4}.$$

(b) Mean and variance

Mean:

$$E(X)=\int_0^2 x\cdot\tfrac34 x(2-x)\,dx=\tfrac34\int_0^2(2x^2-x^3)\,dx=\tfrac34\Big[\tfrac{2x^3}{3}-\tfrac{x^4}{4}\Big]_0^2=\tfrac34\Big(\tfrac{16}{3}-4\Big)=\tfrac34\cdot\tfrac43=1.$$

(By symmetry of $x(2-x)$ about $x=1$, $E(X)=1$ as expected.)

$E(X^2)$:

$$E(X^2)=\tfrac34\int_0^2(2x^3-x^4)\,dx=\tfrac34\Big[\tfrac{x^4}{2}-\tfrac{x^5}{5}\Big]_0^2=\tfrac34\Big(8-\tfrac{32}{5}\Big)=\tfrac34\cdot\tfrac{8}{5}=\tfrac{6}{5}=1.2.$$

Variance:

$$\text{Var}(X)=E(X^2)-[E(X)]^2=1.2-1=0.2=\tfrac15.$$

(c) P(X > 1)

By symmetry about $x=1$, $P(X>1)=\tfrac12$. Verifying:

$$P(X>1)=\tfrac34\int_1^2(2x-x^2)\,dx=\tfrac34\Big[x^2-\tfrac{x^3}{3}\Big]_1^2=\tfrac34\Big[(4-\tfrac83)-(1-\tfrac13)\Big]=\tfrac34\cdot\tfrac23=0.5.$$

(a) Parameter vs statistic; sampling distribution; standard error

• A parameter is a numerical characteristic of the population (e.g. population mean $\mu$, proportion $p$); it is usually fixed and unknown.
• A statistic is a numerical characteristic computed from a sample (e.g. sample mean $\bar x$, sample proportion $\hat p$); it varies from sample to sample and is used to estimate the parameter.
• The sampling distribution is the probability distribution of a statistic over all possible samples of a given size $n$ from the population. Example: the distribution of $\bar x$ from repeated samples of size 36.
• The standard error is the standard deviation of that sampling distribution. For the mean, $\text{SE}(\bar x)=\sigma/\sqrt n$; it measures the typical sampling fluctuation of the estimate and decreases as $n$ increases.

(b) Central Limit Theorem

Statement: If $X_1,X_2,\dots,X_n$ are independent, identically distributed random variables with mean $\mu$ and finite variance $\sigma^2$, then for large $n$ the sampling distribution of the sample mean $\bar X$ is approximately normal:

$$\bar X \;\approx\; N\!\left(\mu,\ \frac{\sigma^2}{n}\right),\quad\text{equivalently}\quad Z=\frac{\bar X-\mu}{\sigma/\sqrt n}\to N(0,1).$$

Significance: It holds regardless of the shape of the population distribution (as long as variance is finite). This justifies using normal-based $z$-tests and confidence intervals for means and proportions in large samples, even when the underlying data are non-normal — making it the foundation of large-sample statistical inference.

Large-sample test of a single proportion. $n=600$, observed $\hat p = 360/600 = 0.60$, claimed $p_0=0.50$.

Hypotheses (two-tailed):

$$H_0: p = 0.50 \qquad H_1: p \ne 0.50.$$

Test statistic (standard normal $z$):

$$z=\frac{\hat p - p_0}{\sqrt{p_0(1-p_0)/n}}=\frac{0.60-0.50}{\sqrt{0.5\times 0.5/600}}=\frac{0.10}{\sqrt{0.0004167}}=\frac{0.10}{0.02041}=4.90.$$

Critical value: at $\alpha=0.01$ two-tailed, $z_{0.005}=\pm 2.576$.

Since $|z|=4.90 > 2.576$, the statistic falls in the rejection region (p-value $\approx 10^{-6}$, far below 0.01).

Conclusion: Reject $H_0$. There is highly significant evidence at the 1% level that the proportion of users preferring Android is not 50% — in fact it is significantly greater than 50% (about 60%).

Chi-square goodness-of-fit test for a fair die.

Hypotheses: $H_0$: the die is fair (each face equally likely, $p_i=1/6$); $H_1$: the die is not fair.

Total rolls $=120$, so expected frequency $E_i = 120/6 = 20$ for every face.

Test statistic:

$$\chi^2=\sum \frac{(O_i-E_i)^2}{E_i}=2.50.$$

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

Since $\chi^2 = 2.50 < 11.07$, we do not reject $H_0$.

Conclusion: The observed frequencies are consistent with a uniform distribution; there is no significant evidence at the 5% level that the die is biased — the die may be regarded as fair.

(Answer any two.)

(a) Skewness and kurtosis

Skewness measures the asymmetry of a distribution about its mean.

• Symmetric: skewness $=0$ (mean = median = mode).
• Positive (right) skew: long right tail, mean > median > mode.
• Negative (left) skew: long left tail, mean < median < mode. A common measure is Karl Pearson's $S_k=(\bar x-\text{mode})/\sigma$.

Kurtosis measures the peakedness and tail-heaviness relative to the normal curve, via $\beta_2=\mu_4/\mu_2^2$.

• $\beta_2=3$: mesokurtic (normal).
• $\beta_2>3$: leptokurtic (sharp peak, heavy tails).
• $\beta_2<3$: platykurtic (flat peak, light tails).

(b) Mathematical expectation (discrete RV)

For a discrete random variable $X$ with pmf $P(X=x_i)=p_i$, the expectation (mean) is

$$E(X)=\sum_i x_i\,p_i.$$

It is the long-run average value of $X$. Properties:

1. $E(c)=c$ for a constant $c$.
2. $E(aX+b)=a\,E(X)+b$ (linearity).
3. $E(X+Y)=E(X)+E(Y)$ (additivity, always).
4. If $X,Y$ are independent, $E(XY)=E(X)E(Y)$.
5. $\text{Var}(X)=E(X^2)-[E(X)]^2$.

(c) Mutually exclusive vs independent events

• Mutually exclusive events cannot occur together: $A\cap B=\varnothing$, so $P(A\cap B)=0$ and $P(A\cup B)=P(A)+P(B)$. Example: rolling a 2 and rolling a 5 on a single die throw.
• Independent events: the occurrence of one does not affect the other, so $P(A\cap B)=P(A)P(B)$. Example: getting a head on a coin toss and a 6 on a die roll.

Note: two events with non-zero probabilities cannot be both mutually exclusive and independent — if mutually exclusive, knowing one occurred tells you the other did not.