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Statistics and Probability are the possibilities of any random occurrence. This phrase refers to determining the likelihood that any given occurrence can occur.
- What are the chances of receiving a head, for instance, when we toss a coin in the air?
- The study of data gathering, analysis, interpretation, presentation, and organization is known as statistics.
- It is a technique for gathering and analyzing data.
- This has several uses on both a local and large scale.
- Statistics are utilized in many data research entities as well, whether it is for studying a nation's population or its economy.
Probability and Statistics
MCQs on Statistics and Probability
Ques. The median of the given data: 17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28 is:
- 10
- 24
- 12
- 8
Click here for Answer
Ans. c) 12
Explanation: List the information in ascending order: 2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48
Since there are an equal number of observations shown here, the average of two middle phrases will be the median.
N / 2th = 16 / 2th = 8th term, where N is the number of terms.
(N / 2 + 1) th = (16 / 2 + 1) th= 9th term
Therefore,
Median = (10 + 14) / 2 = 12
Ques. The probability of each outcome when a coin is tossed 1000 times with the different frequency are: Head:455 & Tail:545
- 0.455 & 0.545
- 0.5 & 0.5
- 0.45 & 0.55
- 455 & 545
Click here for Answer
Ans. a) 0.455 & 0.545
Explanation: Assume that the Head and Tail are occurring at events E and F, respectively.
Probability of Head Occurrence P(E) = Total Number of Heads / Total number of Trials
P(E) = 455 / 1000 = 0.455
Similarly,
P(F) = Total Number of Tails / Total number of Trials
P(F) = 545 / 1000 = 0.545
Ques. A well-shuffled 52 cards deck is used to draw a card. What is the probability of getting a red suit king?
- 3 / 36
- 1 / 26
- 3 / 26
- 1 / 16
Click here for Answer
Ans. b) 1 / 26
Explanation: There are a total of 4 king cards in a 52 cards deck, 2 of which are red and 2 of which are black. Consequently, there are two king cards in the red suit. As a result, the probability of getting a king of red suits is 2 / 52, or 1 / 26.
Ques. From the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, determine the probability that a given number is a multiple of 4.
- 1 / 5
- 1 / 3
- 4 / 12
- 2 / 15
Click here for Answer
Ans. a) 1 / 5
Explanation: S = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.
From the sample space, multiples of 4 are {4, 8, 12}. As a result, 3 / 15 = 1 / 5 represents the probability that the selected number is a multiple of 5.
Ques. What is the probability of landing exactly two tails when two coins are tossed simultaneously?
- 1 / 5
- 1 / 2
- 1 / 3
- None of the above
Click here for Answer
Ans. d) None of the above
Explanation: The sample space, S = {HH, HT, TH, TT}, if two coins are tossed. Successful outcome (getting exactly two tails) = {TT}
Hence, the probability of getting two heads exactly is = 1 / 4.
Ques. Each of a frequency distribution's five continuous classes has a width of 5, and the lowest class's lower class-limit is 10. The highest class's upper class-limit is:
- 15
- 25
- 35
- 40
Click here for Answer
Ans. c) 35
Explanation: The answer to the query is, Since the average breadth of five successive classes is 5, Additionally, the lowest class's upper maximum is 10
We obtain the following information:
10-15
15-20
20-25
25-30
30-35
Thus, we arrive at the highest class's top limit of 35.
Ques. In a continuous frequency distribution, let l be the upper-class limit and m the midpoint. The class's lower-class limit is:
- 2m + l
- 2m – l
- m – l
- m – 2l
Click here for Answer
Ans. b) 2m – l
Explanation: Let x be the continuous frequency distribution's lower-class limit. Let y be the continuous frequency distribution's upper-class limit.
Now, in response to the query, A class's midpoint equals (x + y) / 2 = m
x + y = 2m or x + l = 2m
Given this information, upper-class limit y = l
x = 2m - l
Consequently, the class's lower-class limit is 2m - l.
Read Also: Mean Deviation Continuous Frequency Distribution
Ques. The following is a list of a frequency distribution's class marks: 15, 20, 25, …
The following class is equivalent to the class mark 20:
- 12.5 – 17.5
- 17.5 – 22.5
- 18.5 – 21.5
- 9.5 – 20.5
Click here for Answer
Ans. b) 17.5 – 22.5
Explanation: The answer to the query is, the difference in grades is 5
The following are the classes for these class marks:
2.5 – 7.5
7.5 – 12.5
12.5 – 17.5
17.5 – 22.5
22.5 – 27.5
Consequently, the class equal to class mark 20 = 17.5 - 22.5
Also Read:
| Chapter-Related Topics | ||
|---|---|---|
| Chi-Square Formula | T-Test Formula | Standard Error Formula |
| Skewness Formula | Median | Coefficient of Variation Formula |
Ques. For the following data, a grouped frequency table containing class intervals of equal sizes is created using the range 250 – 270 (270 is not included in this interval).
268, 220, 368, 258, 242, 310, 272, 342,
310, 290, 300, 320, 319, 304, 402, 318,
406, 292, 354, 278, 210, 240, 330, 316,
406, 215, 258, 236.
The frequency for the 310 - 330 range is:
- 4
- 5
- 6
- 7
Click here for Answer
Ans. c) 6
Explanation: When we put the information below in a table, we get,
| Class Interval | Frequency |
|---|---|
| 210 – 230 | 3 |
| 230 – 250 | 3 |
| 250 – 270 | 3 |
| 270 – 290 | 2 |
| 290 – 310 | 4 |
| 310 – 330 | 6 |
| 330 – 350 | 2 |
| 350 – 370 | 2 |
| 370 – 390 | 0 |
| 390 – 410 | 3 |
As a result of the preceding table, we can say that the frequency of the range 310 – 330 is equal to 6.
Ques. For the following data, a grouped frequency distribution table with the classes of similar sizes and 63 – 72 (72 inclusive) as one of the classes is created:
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88,
40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96,
102, 110, 88, 74, 112, 14, 34, 44.
The distribution will have the following number of classes:
- 9
- 10
- 11
- 12
Click here for Answer
Ans. b) 10
Explanation: The answer to the query is, the range of the frequency is between 14 and 112.
As a result, these are the class intervals:
13 - 22, 23 - 32, 33 - 42, 43 - 52, 53 - 62, 63 - 72, 73 - 82, 83 - 92, 93 - 102, 103 - 112.
There are 10 class intervals.
Ques. To express the following frequency distribution with a histogram:
| Class Interval | 5 - 10 | 10 - 15 | 15 - 25 | 25 - 45 | 45 - 75 |
| Frequency | 6 | 12 | 10 | 8 | 15 |
The class 25 – 45's adjusted frequency is:
- 6
- 5
- 3
- 2
Click here for Answer
Ans. d) 2
Explanation: The class's modified frequency is provided by,
(Frequency of given class * Minimum class size of frequency distribution) / Class size of given class
Therefore,
Class 25 – 45 adjusted frequencies = (5 * 8) / 20 = 2.
Also Read:
| Chapter-Related Articles | ||
|---|---|---|
| Experimental Probability | Elementary Event | Sum of Probabilities |
| Geometric Probability | Sure Event | Probability Impossible Events |
Ques. 30 is the mean of the five numbers. Their mean becomes 28 if one of the numbers is left out. The omitted quantity is:
- 28
- 30
- 35
- 38
Click here for Answer
Ans. d) 38
Explanation: Allowing five numbers, a, b, c, d, and e
Data average = sum of all observations / total observations
The answer to the query is
Mean = 30
(a + b + c + d + e) / 5 = 30
⇒ (a + b + c + d + e) = 150 … (1)
Let the excluded number be a
Then, (b + c + d + e) / 4 = 28 as the new mean
⇒ (b + c + d + e) = 112
Equation with this changed (1),
⇒a + 112 = 150
⇒ a = 150 – 112 = 38
So, the excluded number is 38.
Ques. If the observations' mean: x, x + 3, x + 5, x + 7, x + 10
Is 9, the average of the most recent three observations are
- 10 / 3
- 20 / 3
- 30 / 3
- 40 / 3
Click here for Answer
Ans. c) 40 / 3
Explanation: Having said that,
Data average = sum of all observations / total observations
In light of the query
(x + x + 3 + x + 5 + x + 7 + x + 10) / 5 = 9
(5x + 15) / 5 = 9
x + 3 = 9
x = 6
The terms are now
6, 6 + 3, 6 + 5, 6 + 7, 6 + 10 = 6, 9, 11, 13, 16
Therefore, (11 + 13 + 16) / 3 is the last three observations' mean.
= 40 / 3
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