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Question 2 MCQ · 0.0 marks
6406536957219.
Useful Data has been mentioned above.
6406536957220.
This data attachment is just for a reference & not for an evaluation.
Published solution
This is a reference-data card rather than a scored mathematical question. It supplies standard-normal probabilities: \(P(Z<1.645)=0.95\), \(P(Z<1.96)=0.975\), and \(P(Z<-1.645)=0.05\). These are cumulative areas to the left of the listed z-scores and can be used in later normal-distribution questions.
**Answer: A.**
A: **Correct:** Acknowledges the useful reference data, as marked in the source.
B: **Incorrect:** It correctly notes the reference purpose, but is not the source's selected acknowledgement; there is no substantive statistical contrast between these administrative choices.
Match the data types listed in Column A with the corresponding category in Column B, as shown in
Table 1.
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(P)-(3), (Q)-(4), (R)-(2), (S)-(1)
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(P)-(4), (Q)-(3), (R)-(2), (S)-(1)
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(P)-(3), (Q)-(4), (R)-(1), (S)-(2)
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(P)-(1), (Q)-(2), (R)-(3), (S)-(4)
Published solution
Hourly temperatures in one city form a time series because observations are indexed over time. Household incomes across different households in 2025 are cross-sectional because units are compared within one period. Student IDs are nominal labels even though written as numbers. Song lyrics are unstructured text. Thus P–3, Q–4, R–2, S–1.
**Answer: A.**
A: **Correct:** All four mappings match the data's structure or role.
B: **Incorrect:** Interchanges time series and cross-sectional data.
C: **Incorrect:** Student IDs are nominal and song lyrics unstructured; these two are reversed.
D: **Incorrect:** Does not match the categories of any of these examples.
At a school function, it is noticed that 9% of the students are not wearing polished shoes and 14%
are not wearing school ties. These two violations are independent. If a teacher selects 5 students
at random, what is the probability that exactly three of them will be punished for at least one
violation?
Let A mean unpolished shoes and B mean no tie. Independence gives \(P(A\cap B)=0.09(0.14)=0.0126\). A student is punished for either violation, so \(p=P(A\cup B)=0.09+0.14-0.0126=0.2174\). Under the intended independent-student/binomial sampling model, exactly three of five have probability \[\binom53(0.2174)^3(1-0.2174)^2=\binom53(0.2174)^3(0.7826)^2.\] The binomial coefficient chooses which three students are punished. Independence of the two violations alone would not establish independence across students; that is the standard sampling assumption here.
**Answer: B.**
A: **Incorrect:** Uses 0.23, which double-counts students with both violations.
B: **Correct:** Uses the correct union probability and its complement in the binomial formula.
C: **Incorrect:** Adds two separate binomial probabilities and does not count exactly three students with either violation correctly.
D: **Incorrect:** Uses the probability of both violations instead of at least one.
The waiting time for a bus at a certain stop is exponentially distributed with a mean waiting time
of 12 minutes. If there is a 75% chance that a passenger will wait at most
For an exponential waiting time with mean 12, the rate is \(\lambda=1/12\). Its CDF is \(P(T\le t)=1-e^{-t/12}\). Set this equal to 0.75: \(e^{-t/12}=0.25\). Taking natural logarithms gives \(t=-12\ln(0.25)=12\ln4\approx16.64\) minutes.
**Answer: D.**
A: **Incorrect:** 12 ln(0.25) is negative and cannot be this waiting-time percentile.
B: **Incorrect:** Has both the wrong scale and a negative value.
C: **Incorrect:** Uses rate instead of mean as the scale multiplier.
D: **Correct:** 12 ln 4 is the positive 75th percentile.
Standardise the normal variable: \(Z=(X-\mu)/\sigma\). Then \(P(X<\mu+c)=P(Z<c/\sigma)=0.95\). The supplied normal table gives the 95th percentile as 1.645. Thus \(c/\sigma=1.645\), so \(c=1.645\sigma\). A probability such as 0.95 is not itself a z-score.
**Answer: B.**
A: **Incorrect:** Confuses the probability 0.95 with its normal quantile.
B: **Correct:** Uses the correct one-sided 95th percentile and standard-deviation scale.
C: **Incorrect:** 1.96 is the 97.5th percentile, not the 95th.
D: **Incorrect:** Uses the wrong percentile and variance rather than standard deviation.
Rohit wants to open his door with 5 keys(out of which 1 will open the door) and tries the keys
independently and at random. If unsuccessful keys are eliminated from further selection, then find
the expected number of trials required to open the door.
Published solution
Since unsuccessful keys are removed, the trials are not independent with replacement. Equivalently, place the five keys in a random order. The successful key is equally likely to occupy positions 1, 2, 3, 4 or 5. Thus \[E[N]=\sum_{k=1}^5 k\frac15=\frac{1+2+3+4+5}{5}=3.\] A geometric mean of 5 would apply to repeated independent choices with replacement, which is not this procedure.
**Answer: 3.**
The number of system failures in a data center follows a Poisson distribution with an average rate
of 1.2 failures per day. What is the probability that there will be no system failures during a specific
12-hour shift? Enter the answer correct to two decimal places.
Published solution
A 12-hour shift is half a day, so its Poisson mean is \(\lambda=1.2(1/2)=0.6\). For a Poisson variable, \(P(N=k)=e^{-\lambda}\lambda^k/k!\). Therefore \(P(N=0)=e^{-0.6}\approx0.548812\), which rounds to 0.55. Scale the rate to the shift before applying the formula.
**Answer: 0.55.**
The time students spend studying and their scores on an exam are shown in Table 2.
Which of the following option(s) is/are correct?
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There is a strong positive correlation between Study Hours and Exam Score.
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Study hours and Exam Score have a weak correlation.
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The data show that students who study less tend to score higher on the exam.
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The Pearson correlation coefficient (
) for this data set is
.
Published solution
For study hours \(x=(3,7,6,4)\) and scores \(y=(46,55,53,50)\), the means are \(\bar x=5\), \(\bar y=51\). The sums are \(\sum(x-\bar x)(y-\bar y)=20\), \(\sum(x-\bar x)^2=10\), and \(\sum(y-\bar y)^2=46\). Hence Pearson's coefficient is \[r=\frac{20}{\sqrt{10\cdot46}}\approx0.97913\approx0.98.\] The association is strongly positive. Correlation describes these data; it does not alone establish a causal effect of studying.
**Answer: A, D.**
A: **Correct:** r is close to +1, indicating strong positive association.
B: **Incorrect:** A coefficient near 0.98 is not weak.
C: **Incorrect:** The observed trend is increasing, not decreasing.
D: **Correct:** The calculated coefficient rounds to 0.98.
be two random variables. Choose all the correct statements from the following:
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.
6406536957239.6406536957240.
The variance of a Bernoulli random variable is maximum when
, where
is the probability of success.
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=
+
for any two random variables
and
.
Published solution
Use the identities \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\) and \(\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)\). Thus adding 36 does not affect variance and multiplying by 100 multiplies it by 10000. For Bernoulli(p), \(p(1-p)=1/4-(p-1/2)^2\), maximised at p=0.5. In general, \(\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)\); the covariance term cannot always be discarded.
**Answer: B, C.**
A: **Incorrect:** The second-moment identity uses subtraction, not addition.
B: **Correct:** 100 squared is 10000 and a constant shift has no effect.
C: **Correct:** The completed-square expression has its maximum at p=0.5.
D: **Incorrect:** Fails for correlated variables; for example Y=X with positive variance gives 4 Var(X), not 2 Var(X).
The number of students who use different modes of transport to get to school is shown in Figure
1.
Find the relative frequency of students going on a cycle. Enter the answer correct to one decimal
place.
Published solution
The chart shows 5 walking, 10 using the school bus, 6 cycling and 9 using a private van. Total students \(=5+10+6+9=30\). Relative frequency is category count divided by total: \(6/30=0.2\). This is a proportion, equivalent to 20%, and is already correct to one decimal place.
**Answer: 0.2.**
A company has two departments. The Sales department has 30 employees with an average
monthly salary of Rs. 50,000, while the Marketing department has 20 employees with an average
monthly salary of Rs. 60,000. If the company considers both departments together, what is the
overall average monthly salary of all employees?
Published solution
Use a weighted mean because department sizes differ. Sales payroll is \(30\times50000=1500000\), and Marketing payroll is \(20\times60000=1200000\). Combined payroll is 2700000 for 50 employees. Therefore \[\bar x=\frac{30(50000)+20(60000)}{30+20}=54000.\] Simply averaging the two department means would incorrectly weight each department equally.
**Answer: 54000.**
How many three digit numbers are there that do not contain
?
Published solution
The forbidden digit is 4. The hundreds place can use 1–9 except 4, giving 8 choices; zero cannot start a three-digit number. Each of the tens and units places can use 0–9 except 4, giving 9 choices each. Repetitions are allowed because no restriction forbids them. By the product rule, \(8\times9\times9=648\).
**Answer: 648.**
In a program, every student belongs to exactly one of three levels: Foundation, Diploma, or
Degree. A student is chosen at random from the program.
• The probability that the chosen student is at the Foundation or Diploma level is 0.8.
• The probability that the chosen student is at the Diploma or Degree level is 0.6.
Find the probability that the chosen student is at the Diploma level. Enter the answer correct to
one decimal place.
Published solution
Let F, D and G denote the three mutually exclusive levels. Since they exhaust all students, \(P(F)+P(D)+P(G)=1\). Adding the two given probabilities yields \(P(F)+2P(D)+P(G)=0.8+0.6=1.4\). Subtracting 1 gives \(P(D)=0.4\). Equivalently, the two unions cover everyone and overlap exactly in D.
**Answer: 0.4.**
The cumulative distribution function (CDF) of the number of customers (
) waiting in a queue at a
specific time is given as follows:
Based on the above data, answer the given subquestions.
Find the probability that there are exactly 2 customers in the queue. Enter the answer correct to
two decimal places.
Published solution
For a discrete variable, the probability at a value is the jump in its CDF: \(P(X=a)=F(a)-F(a^-)\). At 2 the CDF jumps from 0.60 to 0.85. Hence \(P(X=2)=0.85-0.60=0.25\). The CDF value 0.85 includes all outcomes up to 2, not only 2.
**Answer: 0.25.**
The cumulative distribution function (CDF) of the number of customers (
) waiting in a queue at a
specific time is given as follows:
Based on the above data, answer the given subquestions.
Find the probability that there are at least 1 customer in the queue. Enter the answer correct to
two decimal places.
Published solution
Use the complement: at least one customer means not zero customers. The CDF is 0.25 on \(0\le x<1\), so \(P(X=0)=0.25\). Therefore \(P(X\ge1)=1-P(X=0)=1-0.25=0.75\). Do not use \(1-F(1)\), which would exclude the probability at 1 as well.
**Answer: 0.75.**
