PYQ on Probability and Statistics | Engineering Mathematics

In examinations, questions on Probability and Statistics are considered very important. Most PYQs are based on basic definitions, formulas, and standard theorems related to probability distributions, measures of central tendency, and dispersion.

To solve such questions effectively, one must have a clear understanding of fundamental concepts such as random variables, mean, median, mode, variance, and probability laws, along with the ability to apply these concepts to real-life and theoretical problems.

Short Question on Probability and Statistics

Question 1: A card is drawn at random from a well-shuffled pack of 52 cards. If a king appears, you win ₹50; if a queen, ₹40; for any other card, you lose ₹10. Find your expected gain.

Question 2: A diagnostic test for a disease is 95% accurate. 0.5% of the population has the disease. Find the probability that a person who tests positive actually has the disease.

Question 3: The random variable X can take 0, 1, 2, 3 with probabilities 0.1, 0.2, 0.4, 0.3, respectively. Find the mean, variance, and standard deviation of X.

Question 4: The average marks of students in a class are 70 with SD = 12. If 36 students are chosen at random, find the probability that their average marks are between 68 and 73.

Question 5: Manufacturer claims battery life mean = 400 hr, σ = 50. Sample n = 100 has \bar{x} = 390. Test α = 0.05 whether the mean < 400.

Question 6: Voltages (V) measured: 220, 225, 230, 235, 240. Find mean, median, and SD.

Question 7: Two routers fail independently with probabilities 0.03 and 0.05. Given that at least one failed, probability it was the first router?

Question 8: Raju paid ₹5 to play. Roll a fair die. If you get 6, you win ₹12; if you get 4 or 5, you win ₹4; otherwise, you win nothing. Find your expected gain.

Question 9: A random sample of 36 bulbs shows an average lifetime of 1200 hours with a standard deviation of 90 hours. Find the test statistic and state your decision at the 5% level for testing.

Question 10: The weights of sugar packets are normally distributed with mean μ = 2 kg and standard deviation σ = 0.05 kg. A sample of n = 16 packets is selected. Find the probability that the average weight is between 1.98 kg and 2.02 kg.

Long Type Questions on Probability and Statistics

Question 11: The weights of packets of rice are normally distributed with a mean of 50 kg and a standard deviation of 2 kg. A sample of 25 packets is selected.

• Find the probability that the average weight of the sample is between 49 kg and 51 kg.
• Find the probability that the sample mean is less than 48.5 kg.

Question 12: A machine fills bottles with soft drink. The company claims that the mean volume is 500 ml. A sample of 36 bottles gives a mean of 495 ml with a standard deviation of 8 ml.

• Test at the 5% significance level whether the mean volume differs from 500 ml.
• State the conclusion clearly.

Check if you were right - full answer with solution below.

Solution 1 :

E(X) = 50\left(\frac{52}{4}\right) + 40\left(\frac{52}{4}\right) - 10\left(\frac{52}{44}\right)

E(X) = \frac{200 + 160 - 440}{52} = \frac{-80}{52} = -\text{₹}1.54

Solution 2:

P(D) = 0.005, P (+∣D)=0.95, P( + ∣Dˉ) = 0.05

P(D \mid +) = \frac{P(+ \mid D) P(D)}{P(+ \mid D) P(D) + P(+ \mid \overline{D}) P(\overline{D})}

= \frac{0.95(0.005)}{0.95(0.005) + 0.05(0.995)} \approx 0.087

Solution 3:

E[X] = 0(0.1) + 1(0.2) + 2(0.4) + 3(0.3) = 1.9
E[X²] = 0+1(0.2)+ 4(0.4)+ 9(0.3) = 4.1
Var(X) = 4.1 − (1.9)²= 0.49, σ=0.7.

Solution 4:

\sigma_{\overline{X}} = 2/√36​=2

z_1 = (68−70)/2 = - 1, z₂ ​= (73−70)/2 = 1.5

P = \Phi(1.5) - \Phi(-1) = 0.9332 - 0.1587 = 0.7745

Solution 5:

Set hypotheses: H₀:μ = 400, H₁ : μ < 400 (left-tailed). Test statistic (z):

z = \frac{\overline{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{390 - 400}{50 / \sqrt{100}} = \frac{-10}{5} = -2.0

z_0.05 ​= −1.645
p-value ≈ 0.0228.

Solution 6:

Mean = (220+225+230+235+240)/5 = 230 V

Median = 230 V

Variance = [(220-230)^2 + (225-230)^2 + … + (240-230)^2]/5
= (100+25+0+25+100)/5 = 50
SD=√50 ≈ 7.071 V

Solution 7:

P(at least one fails) = 0.03 + 0.05 - 0.03*0.05 = 0.0785
P(router1 fails | at least one fails) = 0.03/0.0785 ≈ 0.382

Solution 8:

P(6)=1/6,  P(4 or 5) = 2/6 = 1/3,  P(other) = 1/2

Expected payoff = 12⋅1/6+4⋅1/3+0⋅1/2 = 2 + 1.3333 = 3.3333.

Subtract cost ₹5 → Expected gain = 3.3333 − 5 = −1.6667

Solution 9:

Test statistic:

t = \frac{\bar{x}-\mu_0}{s/\sqrt{n}} =1200−125090/6 = −3.33t

Degrees of freedom: df= 35
Critical t-value (two-tailed, 5%) : ±2.030

Decision: ∣t∣ = 3.33 > 2.03 Reject H₀

Solution 10:

σ_Xˉ​ = 0.0125, z_1​ = −1.6,z_2​ = 1.6

z_1​ = 1.98 - 2/0.0125 = -0.02 / 0.0125 = - 1.6

z₂​ = 2.02 - 2/ 0.0125 = 0.02/0.0125 = 1.6

P(1.98 < \bar{X} < 2.02) = P(−1.6 < Z < 1.6) = 0.890

Solution 11:

μ = 50, σ=2 , n = 25

σ_X = σ/\sqrt{n} ​= 2\5 ​= 0.4

For 49 kg:

z₁ = 49−50/0.4 = −2.5

For 51 kg:

z₂ = 51−50/0.4 = 2.

P(49 < \bar{X} < 51) = P(−2.5 < Z < 2.5) = Φ(2.5) − Φ(−2.5) ≈ 0.9938 − 0.0062 = 0.9876

z =48.5 - 50 / 0.4 = - 3.75

P( \bar{X} < 48.5) = Φ(−3.75) ≈ 0.00009

Solution 12:

• Population mean: μ₀ = 500 ml
• Sample mean: xˉ= 495 ml
• Sample size: n = 36
• Sample standard deviation: s = 8 ml
• Significance level: α = 0.05

Use z - test (since n = 36 > 30) :

z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}

dt²/d²x​ = 3(3x + 4y) + 4(−4x + 3y)

dt²/d²x​ = (9 − 16)x + (12 + 12)y = −7x + 24y

4y = dx/dt ​−3x

y = 1/4​(dx/dt ​−3x)

Substitute this value of y in the above equation:

d²x / dt²​= −7x + 24(1/4​(dx/dt​−3x))

d²x​/dt²= −7x + 6dx/dt​ −18x

d²x/dt²​−6dx/dt​+25x = 0

The auxiliary equation is:

m²−6m+25 = 0

m = 3 ± 4i

x = e^3t(Acos4t + Bsin4t)

4y = dx/dt ​−3x

Substitute the value of x:

y = e^3t(−Asin⁡4t + Bcos⁡4t)y

x = e^3t(Acos4t + Bsin4t) , y = e^3t(−Asin⁡4t + Bcos⁡4t)y