INTRODUCTION TO APPLIED STATISTICS QUIZ 3

DATE : 18th OCTOBER, 2024

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Find the expected value for a discrete random variable defined by
$p(x) = \frac{x}{10}, \text{ where } x = 1, 2, 3, 4.$
A. 1.12
B. 3.00
C. 4.20
D. 2.50

Answer: B. 3.00

Explanation:
The expected value $E(X)$ for a discrete random variable is calculated as:

$E(X) = \sum [x \cdot p(x)]$

Substitute the values:

$x = 1$ : $1 \cdot \frac{1}{10} = 0.1$
$x = 2$ : $2 \cdot \frac{2}{10} = 0.4$
$x = 3$ : $3 \cdot \frac{3}{10} = 0.9$
$x = 4$ : $4 \cdot \frac{4}{10} = 1.6$
Sum:

$E(X) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$

Thus, the expected value is 3.00.

One of the following is not true about discrete random variable. Which one is it?
A. The probabilities are non-negative
B. The sum of all probabilities for the given variable must be 1
C. All probabilities must be between 0 and 1
D. The given values of $x$ must be consecutive

Answer: D. The given values of $x$ must be consecutive

Explanation:
For a discrete random variable:

Probabilities are non-negative (A is true).
Sum of probabilities is 1 (B is true).
Each probability $0 \leq p(x) \leq 1$ (C is true).
Values of $x$ need not be consecutive; they can be any discrete set (e.g., $x = 1, 3, 5$ ) (D is false).

The table below gives the distribution of a random variable $X$ .

Answer: C. 1.465

Explanation:
Step 1: Find $a$
Sum of probabilities must be 1:

$0.21 + 0.12 + a + 2 a + 0.21 + 0.01 = 1$

$0.21 + 0.12 + a + 2a + 0.21 + 0.01 = 1$ $0.55 + 3a = 1 \implies 3a = 0.45 \implies a = 0.15$

Thus, probabilities are:

$x=3$ : $a = 0.15$
$x=4$ : $2a = 0.30$

Step 2: Calculate $E(X)$

$E (X) = \sum [x \cdot P (x)] = 1 (0.21) + 2 (0.12) + 3 (0.15) + 4 (0.30) + 5 (0.21) + 6 (0.01)$

$E(X) = \sum [x \cdot P(x)] = 1(0.21) + 2(0.12) + 3(0.15) + 4(0.30) + 5(0.21) + 6(0.01)$ $= 0.21 + 0.24 + 0.45 + 1.20 + 1.05 + 0.06 = 3.21$

Step 3: Calculate $E(X^2)$

$E(X^2) = \sum [x^2 \cdot P(x)] = 1^2(0.21) + 2^2(0.12) + 3^2(0.15) + 4^2(0.30) + 5^2(0.21) + 6^2(0.01)$ $= 1(0.21) + 4(0.12) + 9(0.15) + 16(0.30) + 25(0.21) + 36(0.01) = 0.21 + 0.48 + 1.35 + 4.80 + 5.25 + 0.36 = 12.45$

Step 4: Variance $\sigma^2$

$\sigma^2 = E(X^2) - [E(X)]^2 = 12.45 - (3.21)^2 = 12.45 - 10.3041 = 2.1459$

Step 5: Standard deviation $\sigma$

$\sigma = \sqrt{2.1459} \approx 1.4647 \approx 1.465$

The random variable $X$ has probability function:

$P(X = x) = \begin{cases} kx & \text{for } x = 1,2,3 \\ k(x + 1) & \text{for } x = 4,5 \end{cases}$

where $k$ is a constant. Find the value of $k$ .
A. $\frac{1}{15}$
B. 0.066667
C. $\frac{1}{17}$
D. 15

Answer: C. $\frac{1}{17}$

Explanation:
Sum of probabilities must be 1:

$P (1) + P (2) + P (3) + P (4) + P (5) = 1$

$P(1) + P(2) + P(3) + P(4) + P(5) = 1$ $k (1) + k (2) + k (3) + k (4 + 1) + k (5 + 1) = k + 2 k + 3 k + 5 k + 6 k = 17 k = 1$

$k(1) + k(2) + k(3) + k(4+1) + k(5+1) = k + 2k + 3k + 5k + 6k = 17k = 1$ $k = \frac{1}{17}$

Find the value of $E(X)$ for the random variable in Question 4.
A. 2.1609
B. 1.472
C. 4.267
D. 3.765

Answer: D. 3.765

Explanation:
Using $k = \frac{1}{17}$ :

$E (X) = \sum [x \cdot P (x)] = 1 \cdot P (1) + 2 \cdot P (2) + 3 \cdot P (3) + 4 \cdot P (4) + 5 \cdot P (5)$

$E(X) = \sum [x \cdot P(x)] = 1 \cdot P(1) + 2 \cdot P(2) + 3 \cdot P(3) + 4 \cdot P(4) + 5 \cdot P(5)$ $= 1 (\frac{1}{17}) + 2 (\frac{2}{17}) + 3 (\frac{3}{17}) + 4 (\frac{5}{17}) + 5 (\frac{6}{17})$

$= 1 \left(\frac{1}{17}\right) + 2 \left(\frac{2}{17}\right) + 3 \left(\frac{3}{17}\right) + 4 \left(\frac{5}{17}\right) + 5 \left(\frac{6}{17}\right)$ $= \frac{1}{17} + \frac{4}{17} + \frac{9}{17} + \frac{20}{17} + \frac{30}{17}$

$= \frac{1 + 4 + 9 + 20 + 30}{17}$

$= \frac{64}{17}$

$\approx 3.7647$

$= \frac{1}{17} + \frac{4}{17} + \frac{9}{17} + \frac{20}{17} + \frac{30}{17} = \frac{1+4+9+20+30}{17} = \frac{64}{17} \approx 3.7647 \approx 3.765$

Find $P(2 < X \leq 4)$ for the random variable in Question 4.
A. 0.5333
B. 0.4706
C. 0.6667
D. 0.4667

Answer: B. 0.4706

Explanation:

$P (2 < X \leq 4) = P (X = 3) + P (X = 4)$

$P(2 < X \leq 4) = P(X=3) + P(X=4)$ $P (3) = k \cdot 3 = \frac{1}{17} \cdot 3 = \frac{3}{17}$

$P(3) = k \cdot 3 = \frac{1}{17} \cdot 3 = \frac{3}{17}$ $P (4) = k \cdot (4 + 1) = \frac{1}{17} \cdot 5 = \frac{5}{17}$

$P(4) = k \cdot (4+1) = \frac{1}{17} \cdot 5 = \frac{5}{17}$ $P (2 < X \leq 4) = \frac{3}{17} + \frac{5}{17} = \frac{8}{17}$

$P(2 < X \leq 4) = \frac{3}{17} + \frac{5}{17} = \frac{8}{17} \approx 0.4706$

A random variable $X \sim \text{B}(12, p)$ . The variance is 1.92. Find possible values of $p$ .
A. 2 and 4
B. 0.5 and 0.5
C. 0.12 and 0.88
D. 0.2 and 0.8

Answer: D. 0.2 and 0.8

Explanation:
For binomial distribution, variance $\sigma^2 = np(1-p)$ :

$12 p (1 - p) = 1.92$

$12p(1-p) = 1.92$ $p (1 - p) = \frac{1.92}{12} = 0.16$

$p(1-p) = \frac{1.92}{12} = 0.16$ $p - p^2 = 0.16 \implies p^2 - p + 0.16 = 0$

Solve quadratic equation:

$p = \frac{1 \pm \sqrt{1 - 4(1)(0.16)}}{2} = \frac{1 \pm \sqrt{0.36}}{2} = \frac{1 \pm 0.6}{2}$

$p = \frac{1.6}{2} = 0.8 \quad \text{or} \quad p = \frac{0.4}{2} = 0.2$

40% of students favor introducing scrubs. 8 students are randomly selected. Find probability exactly 4 are in favor.
A. 0.3455
B. 0.2322
C. 0.5806
D. 0.2508

Answer: B. 0.2322

Explanation:
Binomial distribution: $n=8$ , $p=0.4$ , $k=4$

$P (X = 4) = (\binom{8}{4}) (0.4)^{4} (0.6)^{4}$

$P(X=4) = \binom{8}{4} (0.4)^4 (0.6)^4$ $(\binom{8}{4}) = 70, (0.4)^{4} = 0.0256, (0.6)^{4} = 0.1296$

$\binom{8}{4} = 70, \quad (0.4)^4 = 0.0256, \quad (0.6)^4 = 0.1296$ $P = 70 \cdot 0.0256 \cdot 0.1296 = 70 \cdot 0.00331776 = 0.2322432 \approx 0.2322$

Industrial injuries follow Poisson distribution with mean 0.5. Find probability of more than 2 accidents in a week.
A. 0.90980
B. 0.01439
C. 0.05229
D. 0.98561

Answer: B. 0.01439

Explanation:
Poisson distribution: $\lambda = 0.5$

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

$P(X > 2) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$ $P (X = k) = \frac{e^{- λ} λ^{k}}{k!}$

$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ $P (X = 0) = e^{- 0.5} \frac{(0.5)^{0}}{0!} = e^{- 0.5} \approx 0.6065$

$P(X=0) = e^{-0.5} \frac{(0.5)^0}{0!} = e^{-0.5} \approx 0.6065$ $P (X = 1) = e^{- 0.5} \frac{0.5}{1} \approx 0.6065 \cdot 0.5 = 0.30325$

$P(X=1) = e^{-0.5} \frac{0.5}{1} \approx 0.6065 \cdot 0.5 = 0.30325$ $P (X = 2) = e^{- 0.5} \frac{0.25}{2} \approx 0.6065 \cdot 0.125 = 0.0758125$

$P(X=2) = e^{-0.5} \frac{0.25}{2} \approx 0.6065 \cdot 0.125 = 0.0758125$ $P (X \leq 2) \approx 0.6065 + 0.30325 + 0.0758125 = 0.9855625$

$P(X \leq 2) \approx 0.6065 + 0.30325 + 0.0758125 = 0.9855625$ $P (X > 2) = 1 - 0.9855625$

$P(X > 2) = 1 - 0.9855625 = 0.0144375 \approx 0.01439$

Continuous random variable $X$ has pdf:

$f(x) = \begin{cases} kx & 0 \leq x < 2 \\ k(2x - 1) & 2 \leq x \leq 7 \\ 0 & \text{elsewhere} \end{cases}$

Find $k$ .
A. $\frac{1}{1442}$
B. $\frac{1}{17}$
C. $\frac{1}{2044}$
D. $\frac{1}{42}$

Answer: D. $\frac{1}{42}$

Explanation:
Total integral of pdf is 1:

$\int_{0}^{2} kx dx + \int_{2}^{7} k(2x - 1) dx = 1$

First integral:

$\int_{0}^{2} kx dx = k \left[ \frac{x^2}{2} \right]_{0}^{2} = k \left( \frac{4}{2} - 0 \right) = 2k$

Second integral:

$\int_{2}^{7} k(2x - 1) dx = k \left[ x^2 - x \right]_{2}^{7} = k \left[ (49 - 7) - (4 - 2) \right] = k [42 - 2] = 40k$

Sum:

$2k + 40k = 42k = 1 \implies k = \frac{1}{42}$

Calculate $P(X < 3)$ for the random variable in Question 10.
A. 0.612
B. 0.552
C. 0.143
D. 0.024

Answer: C. 0.143

Explanation:

$P(X < 3) = \int_{0}^{2} f(x) dx + \int_{2}^{3} f(x) dx$

First part ( $0 \leq x < 2$ ):

$\int_{0}^{2} \frac{1}{42} x dx = \frac{1}{42} \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{1}{42} \cdot 2 = \frac{2}{42} = \frac{1}{21}$

Second part ( $2 \leq x \leq 3$ ):

$\int_{2}^{3} \frac{1}{42} (2x - 1) dx = \frac{1}{42} \left[ x^2 - x \right]_{2}^{3} = \frac{1}{42} \left[ (9 - 3) - (4 - 2) \right] = \frac{1}{42} [6 - 2] = \frac{4}{42} = \frac{2}{21}$

Sum:

$\frac{1}{21} + \frac{2}{21} = \frac{3}{21} = \frac{1}{7} \approx 0.1429 \approx 0.143$

Find expected value $E(X)$ for the random variable in Question 10.
A. 5.043
B. 3.852
C. 2.454
D. 4.845

Answer: D. 4.845

Explanation:

$E(X) = \int_{0}^{2} x \cdot \frac{1}{42} x dx + \int_{2}^{7} x \cdot \frac{1}{42} (2x - 1) dx$

First integral:

$\int_{0}^{2} \frac{1}{42} x^2 dx = \frac{1}{42} \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{1}{42} \cdot \frac{8}{3} = \frac{8}{126} = \frac{4}{63}$

Second integral:

$\int_{2}^{7} \frac{1}{42} (2 x^{2} - x) d x = \frac{1}{42} {[\frac{2 x^{3}}{3} - \frac{x^{2}}{2}]}_{2}^{7} = \frac{1}{42} ((\frac{2 (343)}{3} - \frac{49}{2}) - (\frac{2 (8)}{3} - \frac{4}{2})]$

$\int_{2}^{7} \frac{1}{42} (2x^2 - x) dx = \frac{1}{42} \left[ \frac{2x^3}{3} - \frac{x^2}{2} \right]_{2}^{7} = \frac{1}{42} \left( \left( \frac{2(343)}{3} - \frac{49}{2} \right) - \left( \frac{2(8)}{3} - \frac{4}{2} \right) \right]$ $= \frac{1}{42} \left( \left( \frac{686}{3} - 24.5 \right) - \left( \frac{16}{3} - 2 \right) \right) = \frac{1}{42} \left( \frac{686}{3} - \frac{49}{2} - \frac{16}{3} + 2 \right)$

Simplify:

$= \frac{1}{42} \left( \frac{670}{3} - \frac{45}{2} \right) = \frac{1}{42} \left( \frac{1340 - 135}{6} \right) = \frac{1}{42} \cdot \frac{1205}{6} = \frac{1205}{252} \approx 4.7825$

Total $E(X)$ :

$\frac{4}{63} + 4.7825 \approx 0.0635 + 4.7825 = 4.846$

(Detailed calculation yields $\frac{1205}{252} \approx 4.7825$ , but earlier $\frac{4}{63} \approx 0.0635$ was incomplete; full recalculation gives $E(X) = \frac{1221}{252} \approx 4.845$ ).

Birthweight uniformly distributed from 2 kg to 5 kg. Find probability weight $\leq 4$ kg.
A. 0.3333
B. 0.2500
C. 0.1453
D. 0.6667

Answer: D. 0.6667

Explanation:
Uniform distribution: $a=2$ , $b=5$ , pdf $f(x) = \frac{1}{3}$

$P(X \leq 4) = \frac{4 - 2}{5 - 2} = \frac{2}{3} \approx 0.6667$

STI symptoms follow uniform distribution from 25 to 45 days. Write pdf.
A. $f(x) = \begin{cases} \frac{1}{20} x & 25 \leq x \leq 45 \\ 0 & \text{elsewhere} \end{cases}$
B. $f(x) = \begin{cases} \frac{1}{20} & 25 \leq x \leq 45 \\ 0 & \text{elsewhere} \end{cases}$
C. $F(x) = \begin{cases} \frac{1}{20} & 25 \leq x \leq 45 \\ 0 & \text{elsewhere} \end{cases}$
D. None

Answer: B. $f(x) = \begin{cases} \frac{1}{20} & 25 \leq x \leq 45 \\ 0 & \text{elsewhere} \end{cases}$

Explanation:
Uniform distribution: pdf is constant over $[a,b]$ .

$a=25, \, b=45, \, \text{range} = 20, \, f(x) = \frac{1}{b-a} = \frac{1}{20}$

Doctor arrives every 8 minutes. Waiting times follow Poisson distribution. Average waiting time?
A. $\frac{1}{8}$ minutes
B. 16 minutes
C. 64 minutes
D. 8 minutes

Answer: D. 8 minutes

Explanation:
In a Poisson process, the average waiting time (interarrival time) is the inverse of the rate. "Every 8 minutes" implies average waiting time is 8 minutes.

Rotting time exponentially distributed with average 5 days. Find probability rotting < 1 day.
A. 0.3625
B. 0.1813
C. 0.0067
D. 0.8187

Answer: B. 0.1813

Explanation:
Exponential distribution: mean $\mu = 5$ , rate $\lambda = \frac{1}{5} = 0.2$

$P (X < 1) = 1 - e^{- λ x}$

$= 1 - e^{- 0.2 \cdot 1}$

$= 1 - e^{- 0.2}$

$\approx 1 - 0.8187$

$P(X < 1) = 1 - e^{-\lambda x} = 1 - e^{-0.2 \cdot 1} = 1 - e^{-0.2} \approx 1 - 0.8187 = 0.1813$

Butchery refunds if rotting time is in first 8%. Cutoff days?
A. 12.62
B. 0.42
C. 6.31
D. 0.52

Answer: B. 0.42

Explanation:
Find $t$ such that $P(X \leq t) = 0.08$ :

$1 - e^{- λ t} = 0.08 ⟹ e^{- λ t} = 0.92 ⟹ - λ t = \ln (0.92) ⟹ t = - \frac{\ln (0.92)}{λ}$

$1 - e^{-\lambda t} = 0.08 \implies e^{-\lambda t} = 0.92 \implies -\lambda t = \ln(0.92) \implies t = -\frac{\ln(0.92)}{\lambda}$ $\ln (0.92) \approx - 0.08338, λ = 0.2$

$\ln(0.92) \approx -0.08338, \, \lambda = 0.2$ $t = - \frac{- 0.08338}{0.2} = \frac{0.08338}{0.2}$

$= 0.4169$

$t = -\frac{-0.08338}{0.2} = \frac{0.08338}{0.2} = 0.4169 \approx 0.42 \text{ days}$

Time to explain illness exponentially distributed with average 12 minutes. Probability > 10 minutes?
A. 0.0232
B. 0.5542
C. 0.4346
D. 0.3012

Answer: C. 0.4346

Explanation:
Mean $\mu = 12$ , $\lambda = \frac{1}{12}$

$P(X > 10) = e^{-\lambda x} = e^{-\frac{10}{12}} = e^{-\frac{5}{6}} \approx e^{-0.8333} \approx 0.4346$

Properties: two outcomes, complementary, independent trials. Which distribution?
A. Poisson
B. Normal
C. Exponential
D. Bernoulli

Answer: D. Bernoulli

Explanation:
Bernoulli distribution describes a single trial with two complementary outcomes (success/failure).

Formula for standard deviation of any probability distribution?
A. $\sqrt{\sum xP(x)}$

B. $\sqrt{\sum x^2 P(x) - [\sum xP(x)]^2}$

C. $\sqrt{\int_a^b x^2 f(x) dx - (\int_a^b xf(x) dx)^2}$

D. $\sqrt{E(X^2) - (E(X))^2}$

Answer: D. $\sqrt{E(X^2) - (E(X))^2}$

Explanation:
Standard deviation $\sigma = \sqrt{\text{Variance}}$ , and variance = $E(X^2) - [E(X)]^2$ . This applies to any distribution.

Probability mass function for complaints:

$X$	0	1	3	4	6
$P(X)$	$\frac{1}{4}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{12}$	$\frac{1}{6}$
Find $F(4)$ .
A. $\frac{3}{4}$
B. $\frac{1}{12}$
C. $\frac{5}{6}$
D. $\frac{11}{12}$

Answer: C. $\frac{5}{6}$

Explanation:
$F(4) = P(X \leq 4) = P(X=0) + P(X=1) + P(X=3) + P(X=4)$

$= \frac{1}{4} + \frac{1}{6} + \frac{1}{3} + \frac{1}{12} = \frac{3}{12} + \frac{2}{12} + \frac{4}{12} + \frac{1}{12} = \frac{10}{12} = \frac{5}{6}$

Expected number of complaints (same pmf as Q21).
A. 1.2
B. 3.0
C. 2.5
D. 2.25

Answer: C. 2.5

Explanation:

$E (X) = \sum [x \cdot P (x)] = 0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{6} + 3 \cdot \frac{1}{3} + 4 \cdot \frac{1}{12} + 6 \cdot \frac{1}{6}$

$E(X) = \sum [x \cdot P(x)] = 0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{6} + 3 \cdot \frac{1}{3} + 4 \cdot \frac{1}{12} + 6 \cdot \frac{1}{6}$ $= 0 + \frac{1}{6} + 1 + \frac{4}{12} + 1$

$= 0 + 0.1667 + 1 + 0.3333 + 1$

$= 0 + \frac{1}{6} + 1 + \frac{4}{12} + 1 = 0 + 0.1667 + 1 + 0.3333 + 1 = 2.5$

Find $E(7 - 2X)$ (same pmf as Q21).
A. 4.600
B. 1.000
C. 2.000
D. 2.500

Answer: C. 2.000

Explanation:

$E(7 - 2X) = 7 - 2E(X) = 7 - 2(2.5) = 7 - 5 = 2$

Which is not true about continuous random variables?
A. $0 \leq P(x) \leq 1$
B. $\int_a^b f(x) dx = 1$
C. Range of values is in an interval
D. $\sum P(x) = 1$

Answer: D. $\sum P(x) = 1$

Explanation:

A: $P(x)$ for a point is 0, which is in [0,1].
B: Total integral of pdf is 1.
C: Continuous variables take values in intervals.
D: Summing probabilities is for discrete; continuous uses integrals.

What is the standard deviation of this variable $X$ ?

All are continuous distributions except:
A. Normal
B. Poisson
C. Exponential
D. Uniform

Answer: B. Poisson

Explanation:
Poisson is a discrete distribution (counts events). Others are continuous.

@Dr. Microbiota

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