Practice AP Inter 1st Year Maths Study Material Chapter 13 Statistics MCQ to identify your strengths and weak areas.
AP Inter 1st Year Maths Statistics MCQ
Question 1.
The range of the following data (marks scored by 10 students in mathematics) is 15, 20, 31, 62, 13, 6, 41, 86, 21, 74 is
1) 68
2) 56
3) 80
4) 71
Answer:
3) 80
Explanation:
Max = 86; Min = 6
Range = Max – Min = 86 – 6 = 80
Question 2.
In a test match two batsmen A and B scored 116, 2 and 76, 138 in two innings respectively. Whose score is more scattered ?
1) A
2) B
3) Can’t say
4) Equal
Answer:
1) A
Explanation:
A’s mean = \(\frac{(116+2)}{2}\) = 59;
B’s mean = \(\frac{(76+138)}{2}\) = 107
⇒ A : deviation from mean = |116 – 59| = 57, |2 – 59| = 57
⇒ B : deviation from mean = |76 – 107| = 31, |138 – 107| = 31
So, A has larger deviation ⇒ more scattered.
Question 3.
The formula \(\frac{1}{N} \sum_{i=1}^n\) fi|xi – x̄| is used to calculate which of the following dispersion?
1) Mean deviation
2) Standard deviation ( 1 )
3) Range
4) Q.D
Answer:
1) Mean deviation
Explanation:
The formula : This formula calculates the Mean Deviation from the mean.
Question 4.
Which of the following formula is used to measure mean deviation for grouped data about median?
(1) \(\sum_{i=1}^{\mathrm{n}}\)fi|xi – x̄|
(2) \(\frac{1}{N_i} \sum_{i=1}^n\)fi|xi – x̄|
(3) \(\frac{1}{N_i} \sum_{i=1}^n\)fi|xi – M|
(4) \(\frac{1}{N_i} \sum_{i=1}^n\)|xi – M|
Answer:
(3) \(\frac{1}{N_i} \sum_{i=1}^n\)fi|xi – M|
Explanation:
Formula for mean deviation for grouped data about median is
\(\frac{1}{N_i} \sum_{i=1}^n\)fi|xi – M|
Question 5.
Standard deviation of a discrete frequency distribution is meansured by using
(1) \(\frac{1}{N} \sqrt{\sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2}\)
(2) \(\sqrt{\sum_{i=1}^n\left(x_i-\bar{x}\right)^2}\)
(3) \(\sqrt{\frac{1}{N_1} \sum_{1=1}^n f_1\left(x_1-\bar{x}\right)^2}\)
(4) \(\sqrt{\frac{1}{\mathrm{~N}_{\mathrm{i}}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{f}_1\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}\)
Answer:
(3) \(\sqrt{\frac{1}{N_1} \sum_{1=1}^n f_1\left(x_1-\bar{x}\right)^2}\)
Explanation:
Standard deviation formula: \(\sqrt{\frac{1}{N_1} \sum_{1=1}^n f_1\left(x_1-\bar{x}\right)^2}\)
Question 6.
If each observation is multiplied by a constant k (≠ 0), the variance of the resulting observations becomes times the original variance ( 4 )
1) k (remains unchanged)
2) √k
3) k3
4) k2
Answer:
4) k2
Explanation:
New variance = k2 × original variance
Question 7.
The variance of 20 observations is 5. If each observation is added by 3 then the new variance of the resulting observations.
1) 5
2) 8
3) 15
4) 9
Answer:
1) 5
Explanation:
Adding (or) subtracting a constant doesn’t change variance.
Question 8.
Mean of the squares of the deviations from mean is known as ( 4 )
1) Mean deviation
2) Standard deviation
3) Quartile deviation
4) Variance
Answer:
4) Variance
Explanation:
This is the variance by defintion.
Question 9.
Relation between the Mean deviation and the Standard deviation is ( 2 )
1) σ ≤ MD
2) MD ≤ σ
3) σ ≤ \(\sqrt{MD}\)
4) σ2 ≤ \(\sqrt{MD}\)
Answer:
2) MD ≤ σ
Explanation:
The relation between mean deviation (MD) and Standard Deviation is always MD ≤ a
Question 10.
The sum of 10 items is 12 and the sum of their squares is 18. The standard deviation is
(a) \(\frac{4}{5}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{1}{5}\)
Answer:
Mean x̄ = \(\frac{12}{10}\) = 1.2
Variance : \(\frac{1}{10}\) Σx2= \(\frac{18}{10}\) – (1.2)2
= 1.8 – 1.44
= 0.36.
Standard deviation = \(\sqrt{0.36}\) = 0.6 = \(\frac{3}{5}\),