AST-01 Solved Assignment January 2026 | Statistical Techniques | IGNOU B.SC.

Q1. Which of the following statements are true? Give reasons for your answer.

(i)) F-distribution is always used in goodness of fit. (60 words)
(ii)) Measure of central tendency in a data set refers to the extent to which the observations are scattered. (60 words)
(iii)) A process is said to be under assignable cause when the points are above the VCL link of (X,R) chart. (60 words)
(iv)) Simple random sampling is done by using random number tables where the probability of drawing a digit is 0.1. (60 words)
(v)) All time series have a trend. (60 words)

Key Points:

F-distribution: Used for ANOVA and variance comparison, not goodness of fit.
Goodness of fit test: Typically uses the Chi-square (χ²) distribution.
Central tendency: Describes typical data value, not data scattering.
Data dispersion: Measures data scattering (e.g., variance, standard deviation).

Answer: The following statements are evaluated based on principles from Statistical Techniques (AST-01), with reasons provided for their truthfulness or falsity. Each sub-question addresses core statistical concepts ranging from distributions and descriptive statistics to sampling methods and quality control.

Q2(a). Following is the distribution of marks (out of 25) obtained by 10 students in Physics and Mathematics. Draw a scatter diagram for all the 10 students and calculate the correlation between marks of Physics and Mathematics.

Key Points:

A scatter diagram visually represents the relationship between two variables.
Pearson's 'r' quantifies the strength and direction of linear correlation (-1 to +1).
Positive correlation: points on scatter diagram rise from lower-left to upper-right.
Formula for 'r': [NΣXY - (ΣX)(ΣY)] / √[[NΣX² - (ΣX)²][NΣY² - (ΣY)²]].

Answer: Understanding the relationship between two variables, such as marks in Physics and Mathematics, is a fundamental concept in statistics. A scatter diagram provides a visual representation of this relationship, while Pearson's correlation coefficient quantifies its strength and direction. Both tools are crucial for analyzing bivariate data as covered in AST-01. ### Scatter Diagram A scatter diagram is a graphical representation used to display the relationship between two quantitative variables....

Q2(b). Two cards are drawn simultaneously or successively without replacement from a well-shuffled deck of 52 cards. Find the probability distribution of number of aces (x). Give a graphical representation of probability distribution.

Key Points:

Random variable 'X' represents the number of aces drawn (0, 1, or 2).
Total outcomes for drawing 2 cards from 52 is C(52, 2) = 1326.
P(X=0) = (C(4,0) × C(48,2)) / C(52,2) = 1128/1326.
P(X=1) = (C(4,1) × C(48,1)) / C(52,2) = 192/1326.

Answer: In the context of statistical techniques, a random variable quantifies the outcome of a random experiment. For this problem, we are drawing two cards simultaneously or successively without replacement from a well-shuffled deck of 52 cards. We aim to find the probability distribution of the number of aces (x) drawn. Let 'X' be the random variable representing the number of aces drawn. Since we are drawing two cards, X can take on the values 0, 1, or 2. A standard deck of 52 cards contains 4 aces...

Q3(a). There are two samples of 1200 and 900 people drawn from populations respectively, which have 30% and 25% of fair-haired people. Test whether the samples drawn from this population maintain difference or not. (Use a = 0.05)

Key Points:

Null (H₀) and alternative (H₁) hypotheses define the claim tested.
Significance level (α=0.05) determines critical Z-values (±1.96).
Pooled proportion (P_pooled) is calculated from total successes and sample sizes.
Z-test statistic measures difference between sample proportions relative to standard error.

Answer: To test whether there is a significant difference between the proportions of fair-haired people in the two populations from which the samples were drawn, we employ a hypothesis test for the difference between two population proportions. The problem implies that the observed proportions of fair-haired people in the samples are 30% and 25%, respectively. **1. State the Hypotheses** The first step involves defining the null and alternative hypotheses. The null hypothesis (H₀) states that there i...

Q3(b). A bicycle shop sells the following number of bicycles from 1990 to 2000. Compute the first three moving averages of length 3 for the bicycle sales data and place them in line with the corresponding year.

Key Points:

Moving average is a time series smoothing technique to identify trends.
A moving average of length 'L' averages 'L' consecutive data points.
Length 3 moving average sums three sales figures and divides by three.
For odd lengths, the moving average is centered on the middle year.

Answer: As an expert IGNOU tutor for AST-01, I will address the question regarding the computation of moving averages. Moving averages are a fundamental technique in time series analysis, specifically discussed in Block 4: Time Series Analysis, Unit 13: Time Series Analysis of your AST-01 course material. Their primary purpose is to smooth out short-term fluctuations and highlight longer-term trends or cycles in data. For this question, we will calculate the first three moving averages of length 3 for b...

Q4(a). Two samples of 9 and 8 sizes give the sum of squares of deviations from respective means equal to 160 and 91 inches squares. Test whether these samples have been drawn from same normal population or not? (Use a = 0.05)

Key Points:

F-test is used to compare two population variances (σ₁² and σ₂²).
H₀: σ₁² = σ₂² (equal variances); H₁: σ₁² ≠ σ₂² (unequal variances).
Sample variance (s²) = Σ(x-x̄)² / (n-1) is calculated for each sample.
F-statistic is computed as the ratio of the two sample variances (e.g., s₁² / s₂²).

Answer: To test whether two samples have been drawn from the same normal population, we typically need to test two conditions: equality of population variances and equality of population means. However, with the given information (sum of squares of deviations from respective means), we can only perform an F-test for the equality of population variances. If the variances are found to be significantly different, then the samples cannot be from the same normal population. First, we establish the null and ...

Q4(b). In a locality of 18,000 families, a random sample of 840 families was taken. Of these 840 families, 206 families were found to have a monthly income of Rs. 500 or less. Give the confidence interval for the families having income Rs. 500 or less.

Key Points:

Point estimate (p̂) is calculated as sample successes (x) divided by sample size (n).
Assumed 95% confidence level, requiring a Z-value of 1.96 for the calculation.
Standard error of proportion (SE_p̂) formula: √(p̂ * (1 - p̂) / n).
Margin of Error (ME) is Z-value multiplied by the standard error of proportion.

Answer: To construct a confidence interval for the proportion of families having a monthly income of Rs. 500 or less, we will follow the standard procedure for estimating a population proportion. Since the confidence level is not specified in the question, we will assume a 95% confidence level, which is a commonly used standard in statistical analysis, as typically covered in AST-01. First, let's identify the given information and calculate the point estimate for the population proportion. The total sa...

Q5(a). Plating of gold on wrist watches requires a chemical called AiCl3. Concentration of the chemical that is added to the solution is an important factor. The objective is to choose that concentration for which plating is uniform. Three concentration of AiCl3 chosen were 5%, 15% and 20%. These were added in the solution and plating was done and thickness (in units) measured on 5 samples is gives below. Use ANOVA to comment on whether the concentration of AiCl3 gives same result or not. (Use a = 0.05)

Key Points:

ANOVA compares means of three or more groups to assess significant differences.
H₀: All group means are equal; H₁: At least one group mean differs.
Calculations involve Sum of Squares (Total, Treatment, Error) and Mean Squares.
F-statistic is the ratio of Mean Square Treatment to Mean Square Error.

Answer: To determine whether the concentration of AiCl3 significantly affects the plating thickness, we employ a one-way Analysis of Variance (ANOVA). ANOVA is a statistical technique used to compare the means of three or more groups, which is precisely our objective here as we have three different concentrations. First, we state the null and alternative hypotheses. The null hypothesis (H₀) posits that there is no significant difference between the mean plating thicknesses for the three concentrations ...

Q5(b). Consider a population of 11 schools from which a sample of size 3 is to be selected. List all the possible samples by circular systematic sampling.

Key Points:

Circular Systematic Sampling selects units from a circular arrangement of the population.
Population units are numbered 1 to N; sample size is n.
Sampling interval (k) is calculated as N/n.
If k is not an integer, an effective integer step size (`k_step = floor(N/n)`) is used for selection.

Answer: Circular systematic sampling is a method used to select a sample from a finite population where the units are arranged in a circular fashion. This method is particularly useful when the population size (N) is not an exact multiple of the desired sample size (n), ensuring that every unit has an equal chance of being selected and that there are N possible samples. In this problem, we have a population of N = 11 schools, and we need to select a sample of size n = 3. First, we number the population...

Q6(a). Consider the case of tossing two dices. Let A: be the event of getting an odd total B: the event of getting on the first dice, and C: the event of getting a total of seven on two dices. Then

(i)) Find P(A∩C), P(A∩B) and P(B∩C). (Brief)
Independence Check) Check whether A,B and C are independent or not. Give reasons for your answers. (50 words)

Key Points:

The sample space for tossing two dice consists of 36 equally likely outcomes.
The probability of an event E is calculated as P(E) = n(E) / n(S).
Intersection of events (A∩B) denotes that both events A and B occur simultaneously.
Events X and Y are independent if and only if P(X∩Y) = P(X) * P(Y).

Answer: This response addresses a probability problem from the AST-01 course, focusing on calculating probabilities of intersections of events and checking for event independence. The scenario involves tossing two dice and defining specific events based on the outcomes.

Q6(b). It is desired to estimate average annual wool yield per sheep for a herd of 150 at a certain farm house using stratified simple random sampling. Sheeps in the herd are to be grouped into three strata on the basis of first shaving in days. Optimum method sample allocation is said to be used for selecting the overall sample of 25 sheeps from the three strata. Determine approximately optimum strata boundaries using the information on first shaving given below.

Key Points:

Stratified sampling divides populations into homogeneous strata for precise estimates.
Optimum (Neyman) allocation distributes sample sizes proportional to stratum size and standard deviation.
Optimum strata boundaries minimize within-stratum variance of the study variable.
Dalenius-Hodges rule determines approximate boundaries using an auxiliary variable's distribution.

Answer: The problem requires the determination of approximately optimum strata boundaries for estimating the average annual wool yield per sheep from a herd of 150, utilizing stratified simple random sampling. This sampling method involves dividing the population into distinct, non-overlapping subgroups, called strata, to improve the precision of estimates. The characteristic chosen for stratification is "first shaving in days," which is expected to be highly correlated with wool yield. Stratification ...

Q7(a). Find the probability that at most 5 defective fuses will be found in a box of 200, if experience shows that 20% of such fuses are defective.

Key Points:

Binomial distribution models fixed trials, two outcomes, constant success probability.
Normal approximation to Binomial used when np ≥ 5 and n(1-p) ≥ 5.
Mean (μ) for Binomial is np; standard deviation (σ) is √[np(1-p)].
Continuity correction converts discrete P(X ≤ k) to continuous P(X < k+0.5).

Answer: This problem involves finding the probability of a certain number of defective items in a large sample, which is a classic application of probability distributions covered in AST-01, specifically the Binomial and its Normal approximation. First, we identify the parameters of the underlying Binomial distribution. We have a fixed number of trials, n = 200 (fuses in a box). Each trial has two possible outcomes: a fuse is defective or not defective. The probability of a fuse being defective is cons...

Q7(b). The following table gives the hours spent by 50 students in the playground in a week: Calculate the mean median and standard deviation of the hours required by students.

Answer: As the specific table for hours spent by 50 students was not provided, we will construct a plausible grouped frequency distribution table to demonstrate the calculation of mean, median, and standard deviation, as would be required in an IGNOU examination. The following calculations are based on this illustrative table. First, we organize the data into a frequency distribution table and calculate the necessary components for mean, median, and standard deviation. This involves finding midpoints ...

Q7(c). In a television appearance, an MLA from a state asks voters in that state to indicate whether the MLA should vote against or for a particular piece of legislation. The MLA office receives 4100 replies. Since voters from western part tends to be democrats while those from eastern parts tend to e communists, the MLA divides the replies on the basis of their geographical origin. 68% of the replies from the western part are against the legislation and 36% of the replies from the eastern part are against it.

(i)) Did the MLA carry out a stratified sampling? (25 words)
(ii)) Did the MLA carry out a stratified random sampling? (25 words)
(iii)) How would you justify the sample design? (25 words)
(iv)) What improvements would you suggest? (25 words)

Key Points:

Stratified sampling divides population into homogeneous subgroups (strata).
MLA's division by geography is a form of stratification based on political views.
Random sampling requires unbiased selection, each unit having equal selection chance.
Voluntary response sampling is biased; respondents self-select, lacking representativeness.

Answer:

Q8(a). At a call centre, callers have to wait till an operator is ready to take their call. To monitor this process, 5 calls were recorded every hour for the 8-hour working day. The data below shows the waiting time in weeks: Compute the control limits for x and R-charts. Draw the control charts for x and R, and comments.

Key Points:

X-bar and R-charts monitor process average and variability, respectively.
Control limits (UCL, LCL) define the expected range of process variation.
R-chart limits: CL=R̄, UCL=D₄R̄, LCL=D₃R̄ (D₃, D₄ depend on subgroup size n).
X-bar chart limits: CL=X̄̄, UCL=X̄̄+A₂R̄, LCL=X̄̄-A₂R̄ (A₂ depends on n).

Answer: Control charts are fundamental tools in Statistical Process Control (SPC) used to monitor a process over time, distinguishing between common cause (random, inherent) and special cause (assignable) variation. For variable data like waiting times, X-bar and R-charts are employed to monitor the process average and variability, respectively. Since the specific waiting time data was not provided in the question, we will use an illustrative dataset. We assume 8 subgroups (hours), each with 5 observat...

Q8(b). Write three advantage and three disadvantage of using a sampling approach instead of a census approach for studying a characteristic.

Key Points:

Sampling is highly cost-effective and significantly reduces study time.
Sampling is practical for large populations or destructive data collection.
Sampling error is an inherent disadvantage, due to observing only a part.
Potential for bias exists if sampling methods are improperly chosen.

Answer: Sampling offers several advantages over a census approach. Firstly, it is highly **cost-effective**, requiring significantly fewer resources for data collection and processing. Secondly, sampling is remarkably **time-saving**, allowing for quicker completion of studies and more timely dissemination of results. Lastly, it is often more **practicable** for extremely large populations or when data collection involves destructive testing, making a census impossible. However, sampling also presents ...

Q9(a). A shop manufacturer is trying to determine whether or not to market a new type of shop. It chooses a random samples of equivalent size in England and France. It asks each of the people in each sample to try the new soap, and see whether he or she likes it better than other soaps. The results are as follows: Test the hypothesis that there are international difference in the proportion of people who prefer the new soap, using 5% level of significance.

Key Points:

Hypothesis testing determines if observed sample differences reflect true population differences.
Null Hypothesis (H₀: p₁=p₂) assumes no difference; Alternative (H₁: p₁≠p₂) assumes a difference.
A Z-test for two proportions is used to compare proportions from two independent samples.
Pooled proportion (p̄) estimates the common population proportion under the null hypothesis.

Answer: To determine whether there are international differences in the proportion of people who prefer a new type of soap, we conduct a hypothesis test for the difference between two population proportions. This statistical technique is fundamental in courses like AST-01 for comparing characteristics across different groups. We will assume the following illustrative sample data, as specific results were not provided in the problem statement for England and France: • **England (Sample 1):** Sample s...

Q9(b). Suppose from a total of 120 guava trees, 5 clusters of 4 trees each selected and the yield (in kg) is recorded below: Estimate the average yield per tree and its standard error.

Key Points:

Cluster sampling is an efficient method for estimating population parameters from grouped units.
Average yield per tree (ȳ) is estimated by averaging all sampled individual tree yields.
The calculation of average yield can also be done by averaging the mean yields of sampled clusters.
The standard error (SE) quantifies the precision of the estimated average yield.

Answer: The problem requires estimating the average yield per tree and its standard error using cluster sampling. This method is particularly useful when the population is naturally grouped into clusters, making it more cost-effective and practical than simple random sampling. In this scenario, guava trees are grouped into clusters of 4 trees each. Since the yield data was not provided in the question, we will assume a hypothetical set of yield data (in kg) for the 5 selected clusters to demonstrate th...

Q10(a). A manufacture of rayon wants to compare that the yield strength of 11.5kg/mm² is met or not at 5% level of significance. The manufacturer draws a sample and calculates the mean to be 12.8kg/mm² and the standard derivation is known to be 2.0kg/mm². Carry out the statistical test appropriate for this.

Key Points:

Z-test is used for population mean when population standard deviation is known.
Null Hypothesis (H₀): μ = 11.5; Alternative (H₁): μ ≠ 11.5 (two-tailed test).
Level of significance (α) is 0.05, leading to critical Z-values of ±1.96.
Test statistic Z = (x̄ - μ₀) / (σ/√n); sample size 'n' must be assumed if not given.

Answer: To determine whether the yield strength of rayon meets the specified 11.5 kg/mm² at a 5% level of significance, an appropriate statistical test is the Z-test for a population mean. This test is selected because the population standard deviation (σ) is known, and the sample mean is given, allowing us to compare it against a hypothesized population mean. **Step 1: Formulate the Null and Alternative Hypotheses** The null hypothesis (H₀) states that the population mean yield strength (μ) is equal ...

Q10(b). A box contains 10 screws, 3 of which are defective. Two screws are drawn at random. Find the probability that none of them is defective, if the sample is drawn

(i)) with replacement, (Brief)
(ii)) without replacement. (Brief)

Key Points:

Total probability uses favorable outcomes divided by total outcomes.
"With replacement" implies independent events, probabilities multiply directly.
P(A and B) = P(A) * P(B) for independent events.
"Without replacement" implies dependent events, sample space changes.

Answer: This problem from AST-01 (Statistical Techniques) deals with calculating probabilities for successive draws, a fundamental concept in statistics. We are asked to determine the probability of drawing two non-defective screws from a box containing 10 screws (3 defective, 7 non-defective), under two different sampling conditions: with replacement and without replacement. Understanding the distinction between these methods is crucial, as it impacts whether the probability of successive events remain...

Q10(c). A researcher measures the temperature of a solution 5 times. The observations are 25, 32, 27, 29, 30 ° C. Determine an unbiased estimate of the temperature.

Key Points:

An unbiased estimate means the estimator's expected value equals the true parameter.
The sample mean (x̄) is the unbiased estimator for the population mean (μ).
The population mean represents the true underlying temperature of the solution.
Given observations are 25, 32, 27, 29, 30 °C, with n = 5.

Answer: In statistics, particularly in the context of estimation theory discussed in AST-01, the primary goal is often to use sample data to make inferences about unknown population parameters. When we want to determine an unbiased estimate of a population parameter, we select an estimator whose expected value is equal to the true value of the parameter being estimated. This property, known as unbiasedness, ensures that, on average, the estimator does not systematically overestimate or underestimate the...

AST-01 Solved Assignment — January 2026 / July 2026

Statistical Techniques | IGNOU Bachelor of Science (B.SC.)

AST-01—January 2026 / July 2026

Statistical Techniques

Questions & Answers

Download Study Materials

Related Courses

AST-01 Assignment Questions — January 2026

Q1. Which of the following statements are true? Give reasons for your answer.

Q2(a). Following is the distribution of marks (out of 25) obtained by 10 students in Physics and Mathematics. Draw a scatter diagram for all the 10 students and calculate the correlation between marks of Physics and Mathematics.

Q2(b). Two cards are drawn simultaneously or successively without replacement from a well-shuffled deck of 52 cards. Find the probability distribution of number of aces (x). Give a graphical representation of probability distribution.

Q3(a). There are two samples of 1200 and 900 people drawn from populations respectively, which have 30% and 25% of fair-haired people. Test whether the samples drawn from this population maintain difference or not. (Use a = 0.05)

Q3(b). A bicycle shop sells the following number of bicycles from 1990 to 2000. Compute the first three moving averages of length 3 for the bicycle sales data and place them in line with the corresponding year.

Q4(a). Two samples of 9 and 8 sizes give the sum of squares of deviations from respective means equal to 160 and 91 inches squares. Test whether these samples have been drawn from same normal population or not? (Use a = 0.05)

Q4(b). In a locality of 18,000 families, a random sample of 840 families was taken. Of these 840 families, 206 families were found to have a monthly income of Rs. 500 or less. Give the confidence interval for the families having income Rs. 500 or less.

Q5(b). Consider a population of 11 schools from which a sample of size 3 is to be selected. List all the possible samples by circular systematic sampling.

Q6(a). Consider the case of tossing two dices. Let A: be the event of getting an odd total B: the event of getting on the first dice, and C: the event of getting a total of seven on two dices. Then

Q7(a). Find the probability that at most 5 defective fuses will be found in a box of 200, if experience shows that 20% of such fuses are defective.

Q7(b). The following table gives the hours spent by 50 students in the playground in a week: Calculate the mean median and standard deviation of the hours required by students.

Q8(b). Write three advantage and three disadvantage of using a sampling approach instead of a census approach for studying a characteristic.

Q9(b). Suppose from a total of 120 guava trees, 5 clusters of 4 trees each selected and the yield (in kg) is recorded below: Estimate the average yield per tree and its standard error.

Q10(b). A box contains 10 screws, 3 of which are defective. Two screws are drawn at random. Find the probability that none of them is defective, if the sample is drawn

Q10(c). A researcher measures the temperature of a solution 5 times. The observations are 25, 32, 27, 29, 30 ° C. Determine an unbiased estimate of the temperature.

More AST-01 Solved Materials