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Class 12 Applied Maths Chapter 10 Solutions
Inferential Statistics
EXERCISE- 10.2
Q.1 A simple random sample of 50 items from a population with σ = 6 resulted in a sample mean of 32.
(i) Provide a 90% confidence interval for the population mean.
(ii) Provide a 99% confidence interval for the population mean.
Ans.(i) Sample size n = 50
Population standard deviation σ = 6
Sample mean x̄ = 32
Confidence level = 90%, so Zα/2 = 1.645
Margin of Error (E) = Zα/2 x 
= 1.645 x 
= = 1.645 x 
= 1.396
= 1.4
Now, Confidence Interval :
µ = x̄ – E , µ = x̄ + E
= 32 – 1.4 , = 32 + 1.4
= 30.6 , = 33.4
(30.6 , 33.4)
(ii) Sample size n = 50
Population standard deviation σ = 6
Sample mean x̄ = 32
Confidence level = 99%, so Zα/2 = 2.576
Margin of Error (E) = Zα/2 ×
= 2.576 x
= 2.576 x
= 2.186
= 2.19
Now, Confidence Interval :
µ = x̄ – E , µ = x̄ + E
= 32 – 2.19 , = 32 + 2.19
= 29.81 , = 34.19
(28.81 , 34.19)
Q.2 A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is σ = 15.
(i) Compute the 95% confidence interval for the population mean.
(ii) Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean.
Ans.(i) Sample size n = 60
Population standard deviation σ = 15
Sample mean x̄ = 80
Confidence level = 95%, so Zα/2 = 1.960
Margin of Error (E) = Zα/2 ×
= 1.960 x 
= 1.960 x 
= 3.79
Now, Confidence Interval :
µ = x̄ – E , µ = x̄ + E
= 80 – 3.79 , = 80 + 3.79
= 76.21 , = 83.79
(76.2 , 83.8)
(ii) Sample size n = 120
Population standard deviation σ = 15
Sample mean x̄ = 80
Confidence level = 95%, so Zα/2 = 1.960
Margin of Error (E) = Zα/2 ×
= 1.960 x 
= 1.960 x 
= 2.68
Now, Confidence Interval :
µ = x̄ – E , µ = x̄ + E
= 80 – 2.68 , = 80 + 2.68
= 77.32 , = 82.68
(77.32 , 82.68)
Q.3 A survey for the prices of a particular product in different stores of a city is conducted. The prices of the product in 16 different stores are given below:
₹100, ₹108, ₹97, ₹99, ₹112, ₹99,₹98, ₹104, ₹107, ₹95, ₹110, ₹103, ₹110, ₹111, ₹106, ₹105. Assuming that the prices of this product follow a normal law with variance of 25.
(i) Compute the sample mean of the distribution.
(ii) Determine the 95% confidence interval for the population mean.
Ans.(i) Compute the sample mean of the distribution.
x̄ = 
= 104
(ii) σ2 = 25, σ = 5
Confidence level = 95%, so Zα/2 = 1.96
Margin of Error (E) = Zα/2 ×
= 1.960 x 
= 1.960 x 
= 2.45
Now, Confidence Interval :
µ = x̄ – E , µ = x̄ + E
= 104 – 2.45 , = 104 + 2.45
= 101.55 , = 106.45
(101.55 , 1.6.45)
Q.4 The average height of a random sample of 400 adult males of a city is 175 cm. It is known that the population standard deviation is 40.
(i) Determine the 90% confidence interval for the population mean.
(ii) Determine the 95% confidence interval for the population mean.
Ans.(i) Sample size n = 400
Population standard deviation σ = 40
Sample mean x̄ = 175
Confidence level = 90%, so Zα/2 = 1.645
Margin of Error (E) = Zα/2 ×
= 1.645 x 
= 1.645 x 
= 3.290
= 3.3
Now, Confidence Interval :
µ = x̄ – E , µ = x̄ + E
= 175 – 3.3 , = 175 + 3.3
= 171.7 , = 178.3
(171.7 , 178.3)
(ii) Sample size n = 400
Population standard deviation σ = 40
Sample mean x̄ = 175
Confidence level = 95%, so Zα/2 = 1.96
Margin of Error (E) = Zα/2 ×
= 1.96 x
= 1.96 x
= 3.92
Now, Confidence Interval :
µ = x̄ – E , µ = x̄ + E
= 175 – 3.92 , = 175 + 3.92
= 171.08 , = 178.08
(171.08 , 178.08)
Q.5 The monthly sales of a plywood shop are normally distributed with a standard deviation of ₹900. A statistical study of sales in the last nine months has found a confidence interval for the mean of monthly sales with extremes of ₹4663 and ₹5839.
(i) What were the average sales over the nine month period?
(ii) What is the confidence level for this interval?
Ans. (i) To find the average (mean) of the confidence interval, we take the midpoint of the lower and upper bounds:
x̄ = 
= 
= Rs.5251
(ii) Confidence interval = (4663, 5839), Standard deviation σ = 900
Sample size n = 9,
Margin of Error (E) = 5839 – 5251 = ₹588
Margin of Error (E) = Zα/2 ×
588 = Zα/2 x 
Zα/2 = 
= 1.96
Therefore, Confidence level = 95%
Q.6 A 99% confidence interval for a population mean was reported to be 83 to 87. If o = 8, what sample size was used in this study?
Ans. Confidence interval = (83, 87) , Zα/2 = 2.576
Population standard deviation σ = 8
Confidence level = 99%
Sample Mean (n) = ?
x̄ – E = 83………(i)
x̄ + E = 87………(ii)
2x̄ = 170
x̄ = 85
From eq (i), x̄ – E = 83
85 – E = 83
E = 2
Margin of Error (E) = Zα/2 ×
2 = 2.576 x 
= 2.576 x 
= 10.3.4
n = 106.17
Therefore, n = 106
Q.7 A simple random sample of 800 elements generates a sample proportion p̄ = 0.70
(i) Provide a 90% confidence interval for the population proportion.
(ii) Provide a 95% confidence interval for the population proportion.
Ans. Sample size n = 800
Sample proportion p̄ = 0.70
(i) Confidence level = 90%
Margin of Error (E) = Zα/2 x
= 1.645 x 
= 
x 
= 
x 
= 0.1645 x 1.62
= 0.026649 = 0.0267
Now, Confidence Interval :
µ = p̄ – E , µ = p̄ + E
= 0.7 – 0.0267 , = 0.7 + 0.0267
= 0.6733 , = 0.7267
(0.6733 , 0.7318)
(ii) Confidence level = 95%
Margin of Error (E) = Zα/2 x
= 1.96 x
= 
x
= x
= 0.031752
Now, Confidence Interval :
µ = p̄ – E , µ = p̄ + E
= 0.7 – 0.031752 , = 0.7 + 0.031752
= 0.6682 , = 0.7318
(0.6682 , 0.7318)
Q.8 In a survey, the sample proportion for the population proportion is p̄ = 0.35. How large a sample be taken to provide a 95% confidence level with a margin of error of 0.05?
Ans. Sample proportion p̄ = 0.35
Margin of Error E = 0.05
Confidence Level = 95%
Margin of Error (E) = Zα/2 x
0.05 = 1.96 x 
= 1.96 x 
n = (18.6984)2
= 349.6
= 350
Q.9 A survey of 611 office workers investigated telephone answering practices, including how often each office worker was able to answer incoming telephone calls and how often incoming telephone calls went directly to voice mail. A total of 281 office workers indicates that they never need voice mail and are able to take every telephone call.
(i) What is the point estimate of the population proportion of office workers who are able to take every telephone call?
(ii) At 90% confidence level, what is the margin of error?
(iii) What is the 90% confidence interval for the proportion of population of office workers who are able to take every telephone call?
Ans. Sample size n = 611
Favourable responses x = 281
(i) The point estimate is the sample proportion:
p̄ = 
= 0.4599
(ii) Confidence level = 90%, SO Zα/2 = 1.645
Margin of Error (E) = Zα/2 x
= 1.645 x 
= 1.645 x 0.0201
= 0.033
(iii) Now, Confidence Interval :
µ = p̄ – E , µ = p̄ + E
= 0.46 – 0.033 , = 0.460 + 0.033
= 0.427 , = 0.493
Q.10 According to a financial report, the majority of companies reporting profits had beaten estimates. A sample of 162 companies showed 104 beat estimates, 29 matched estimates and 29 fell short.
(i) What is the point estimate of the proportion that fell short of estimates?
(ii) Determine the margin of error and provide a 95% confidence interval for the proportion that beates estimates.
(iii) How large a sample is needed if the desired margin of error is 0.05?
Ans. Sample size n = 60
Number of beat estimates = 104
Number of matched estimates = 29
Number of 29 fell short = 29
(i) The point estimate is the sample proportion:
p̄ = 
= 0.179
(ii) Confidence level = 95%
p̄ = 
= 0.6419
Margin of Error (E) = Zα/2 x
= 1.96 x 
= 1.96 x 0.037
= 0.07252 = 0.073
Now, Confidence Interval :
µ = p̄ – E , µ = p̄ + E
= 0.6419 – 0.073 , = 0.6419 + 0.073
= 0.5689 , = 0.7149
(iii) Margin of Error (E) = 0.05
Margin of Error (E) = Zα/2 x
0.05 = 1.96 x 
Squaring Both the sides, we get
0.0025 = 
n = 
n = 353
FAQ’s related to Class 12 Applied Maths Chapter 10 on Inferential Statistics:
Q1. What is Inferential Statistics?
Ans. Inferential Statistics is the branch of statistics that allows us to make predictions or inferences about a population based on a sample. It involves estimating parameters, testing hypotheses, and drawing conclusions using data analysis.
Q2. How is Inferential Statistics different from Descriptive Statistics?
Ans. (i) Descriptive Statistics deals with summarizing and presenting data (like mean, median, mode).
(ii) Inferential Statistics goes further to draw conclusions about the larger population from a small sample using tools like confidence intervals and hypothesis testing.
Q.3 What are the main concepts covered in this chapter?
Ans. (i) Population and Sample
(ii) Parameters and Statistics
(iii) Sampling methods
(iv) Types of errors (Type I and Type II)
(v) Confidence Interval
(vi) Hypothesis Testing
(vii) Level of Significance
Q.4 What is a population and sample in statistics?
Ans. (i) Population is the entire group you want to study (e.g., all students in India).
(ii) Sample is a subset taken from the population for analysis (e.g., 500 students selected randomly).
These are a few Frequently Asked Questions relating to Class 12 Applied Maths Chapter 10
In Class 12 Applied Maths chapter 10, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 12 Applied Maths Chapter 10 promises an enriching learning experience that will set you on the path to success. Class 12 Applied Maths Chapter 10, we delve deep into advanced mathematical concepts that are crucial for understanding.
