GTU BE03000251 Probability and Statistics Winter 2025 Solved Paper & Solutions

A committee of 5 is chosen from $6 + 9 = 15$ people, so the total number of ways is $\binom{15}{5}$. The favourable ways are 3 men from 6 and 2 women from 9.

Example: the number of heads in three tosses of a coin, which takes the values 0, 1, 2 or 3. The number of defective items in a lot, and the number of accidents in a day, are also discrete.

Example: the lifetime of a bulb in hours, the height of a student, or the current in a wire. The value may be any number such as 100.37 hours.

Main difference: a discrete variable is counted and has jumps, and a continuous variable is measured and takes values without a gap.

Let $T_1, T_2, T_3$ be the events that the flashlight is of type 1, 2 or 3, and let $A$ be the event that it gives more than 100 hours of use.

Given: $P(T_1) = 0.2$, $P(T_2) = 0.3$, $P(T_3) = 0.5$, and $P(A|T_1) = 0.7$, $P(A|T_2) = 0.4$, $P(A|T_3) = 0.3$.

(a) Total probability. The three types cover all flashlights, so by the theorem of total probability

(b) Bayes' theorem. The probability that the flashlight is of type 2, given that it lasted more than 100 hours:

For an exponential distribution with mean $\dfrac{1}{\lambda} = 10$, the rate is $\lambda = 0.1$. The density is $f(x) = 0.1e^{-0.1x}$ for $x > 0$, and $P(X > a) = e^{-\lambda a}$.

(a) $P(X > 10) = e^{-0.1 \times 10} = e^{-1} = 0.3679$

(b) $P(X < 30) = 1 - e^{-0.1 \times 30} = 1 - e^{-3} = 1 - 0.0498 = 0.9502$

Let $X$ be the number of accidents in a day. The accidents are rare and independent events, so $X$ follows a Poisson distribution with $\lambda = 3$, so $P(X = k) = \dfrac{e^{-3}3^k}{k!}$.

Each call is a trial with the probability that the line is occupied, $p = 0.4$, and not occupied, $q = 0.6$. Let $X$ be the number of calls with an occupied line out of $n = 10$. $X$ follows a binomial distribution, $P(X = r) = \binom{10}{r}(0.4)^r(0.6)^{10-r}$.

(a) Exactly 3 occupied:

(b) At least one call with the line not occupied. The complement is that all 10 calls find the line occupied:

(c) Expected number occupied:

Rank the marks (the highest mark is rank 1).

$\sum d^2 = 60$ and $n = 9$. Spearman's rank correlation:

We use Karl Pearson's coefficient of skewness, $Sk = \dfrac{\text{Mean} - \text{Mode}}{\sigma}$. The class width is $h = 10$. Take the assumed mean $A = 115$ and $d = \dfrac{x - 115}{10}$.

Hypotheses: $H_0: \mu = 1000$ against $H_1: \mu > 1000$ (a right tailed test). $\alpha = 0.05$, so the critical value is $z_{0.05} = 1.645$.

Given: $n = 64$, $\bar x = 1038$ and $s = 146$. The sample is large, so use the $z$ test with $s$ in place of $\sigma$.

Decision: $z = 2.08 > 1.645$, so it falls in the rejection region.

Hypotheses: $H_0: \sigma_1^2 = \sigma_2^2$ against $H_1: \sigma_1^2 < \sigma_2^2$. $\alpha = 0.05$.

For this one sided test, the larger sample variance goes in the numerator, and the $F$ ratio is

The degrees of freedom are $(12 - 1, 12 - 1) = (11, 11)$, and the critical value is $F_{0.05}(11, 11) = 2.82$.

Decision: $F = 3.14 > 2.82$, so we reject $H_0$.

Hypotheses: $H_0: \mu_1 - \mu_2 = 0$ against $H_1: \mu_1 - \mu_2 \ne 0$ (a two tailed test), $\alpha = 0.10$. Each tail has 0.05, so the critical values are $z = \pm 1.645$.

Given: $n_1 = n_2 = 50$, $\bar x_1 = 6.8$, $s_1 = 0.85$, $\bar x_2 = 7.3$, $s_2 = 1.2$. The samples are large, so use the $z$ test:

Decision: $|z| = 2.40 > 1.645$, so it falls in the rejection region.

The gamma distribution with shape $\alpha > 0$ and scale $\beta > 0$ has the probability density function

and $f(x) = 0$ otherwise.

Mean: $E(X) = \alpha\beta$.

Variance: $V(X) = \alpha\beta^2$.

When $\alpha = 1$ it becomes the exponential distribution with mean $\beta$. The gamma function is $\Gamma(\alpha) = \int_0^\infty t^{\alpha - 1}e^{-t}\,dt$, and the distribution is used for waiting times, the lifetime of a system and the amount of rain.

The density is $f(x) = \dfrac{x}{8}$ for $3 < x < 5$. We check that it is a valid density: $\displaystyle\int_3^5\frac{x}{8}\,dx = \left[\frac{x^2}{16}\right]_3^5 = \frac{25 - 9}{16} = 1$.

(a)

(b)

Take the natural logarithm of $y = ae^{bx}$:

Put $Y = \ln y$ and $A = \ln a$. Then $Y = A + bx$ is a straight line, which is fitted by least squares with the normal equations

Here $n = 5$ and $\sum x = 50$. The normal equations are

Solve:

Check: at $x = 10$, $y = 97.91e^{1.510} = 444$, which is close to 450.