Problem Solving and Data Analysis MCQ Quiz - Objective Question with Answer for Problem Solving and Data Analysis - Download Free PDF

Solution:

We are given:

60% boys → P(Boy) = 0.6

40% girls → P(Girl) = 0.4

8% of boys got 'A' → P(A | Boy) = 0.08

12% of girls got 'A' → P(A | Girl) = 0.12

We are to find: P(Boy | A)

Use Bayes' Theorem:

P(Boy | A) = [P(A | Boy) × P(Boy)] / P(A)

First, compute total probability of getting 'A':

P(A) = P(A | Boy) × P(Boy) + P(A | Girl) × P(Girl)

      = 0.08 × 0.6 + 0.12 × 0.4

      = 0.048 + 0.048 = 0.096

Now, compute P(Boy | A):

P(Boy | A) = (0.08 × 0.6) / 0.096 = 0.048 / 0.096 = 0.5

The problem-solving method is an instructional approach that engages students in identifying, analyzing, and solving problems through critical thinking and inquiry.

Key Points

• Being a structured and inquiry-based approach, the problem-solving method is generally not a time-saving method.
• In fact, it often requires more time than traditional methods because students explore multiple possibilities, gather information, test solutions, and reflect on their findings.
• The focus is on deep understanding rather than quick coverage of content.
• While it is highly beneficial in promoting meaningful learning, it demands considerable classroom time for planning, execution, and discussion.

Hint

•  Helping in the development of scientific attitude is a key strength, as students learn to think logically, question assumptions, and make evidence-based conclusions.
• Training in the scientific method occurs naturally in this approach since learners go through steps such as problem identification, hypothesis formation, experimentation, and evaluation.
• Being a learner-centered method, it places the student at the core of the learning process, encouraging autonomy, curiosity, and active participation.

Hence, the correct answer is time saving method.

Explanation:

Given:

$P(A)=P(\frac{A}{B})=0.25$

and $P(\frac{B}{A})=0.5$

I. $P(\frac{B}{A})=\frac{P(A\cap B)}{P(A)}$

⇒ P(A∩B) = P(A) P(B|A)

⇒ P(A∩B) = 0.25 × 0.5 = 0.125

Now

⇒ $P(\frac{B}{A})=\frac{P(A\cap B)}{P(B)}$

⇒ $P(B)=\frac{P(A\cap B)}{P(\frac{A}{B})}$

⇒ $P(B)=\frac{0.125}{0.25}=0.5$

Now, P(A).P(B) = 0.25 × 0.5 = 0.125 = P(A∩B)

Thus  A and B are independent

II. $P(\bar{A}\cup \bar{B})=1–P(A\cap B)$

= 1 – 0.125 = 0.875

III. $P(\bar{A}\cap \bar{B})=1–P(A\cup B)$

= 1 – [P(A) + P(B) – P(A ∩ B)

=  1 – [0.25 + 0.5 – 0.125]

= = 1 – 0.625 = 0.375

So all statements I, II, and III are correct.

∴ Option (d) is correct.

CONCEPT:

Mean and Median of a Data Set

• Mean is the average of all data values:

  Mean = (Sum of all values) / (Number of values)

• Median is the middle value when the data is arranged in ascending order.
• For an even number of values, median = average of the two middle terms.

EXPLANATION:

• Given data: {25, 29, 25, 32, 24, x}
• Mean = 27
• Total sum = 6 × 27 = 162
• Sum of known values = 25 + 29 + 25 + 32 + 24 = 135
• So, x = 162 − 135 = 27

Final data set: {24, 25, 25, 27, 29, 32}

• Arrange in ascending order: 24, 25, 25, 27, 29, 32
• Middle two values = 25 and 27
• Median = (25 + 27) / 2 = 26

Correct answer  26

m = x₂−x₁/ y_2​ − y_1​​

Substituting the points (x1,y1)=(2,18)Unknown node type: span" id="MathJax-Element-8-Frame" role="presentation" style="position: relative;" tabindex="0">(x1,y1)=(2,18)Unknown node type: span$(x_1,y_1)=(2,18)\; <merror><mtext>Unknown\; node\; type:\; span</mtext></merror>$ (x1,y1)=(2,18)<span style="font-family: var(--bs-body-font-family); font-size: var(--bs-body-font-size); font-weight: var(--bs-body-font-weight); text-align: var(--bs-body-text-align);"> </span> (x2,y2)=(5,22) (x_2, y_2) = (5, 22)" id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0">(x2,y2)=(5,22)$(x_2, y_2) = (5, 22)$ $(x_2, y_2) = (5, 22)$

 

$$m = \frac{22 - 18}{5 - 2}$$ m=43=1.33

 

The slope m m" id="MathJax-Element-13-Frame" role="presentation" style="position: relative;" tabindex="0">m$m$ $m$  of the best-fit line is 1.33, meaning for every 1 unit increase in X, Y increases by 1.33 units. $$m = \frac{4}{3} = 1.33$$
 

The problem-solving method is an instructional approach that engages students in identifying, analyzing, and solving problems through critical thinking and inquiry.

Key Points

• Being a structured and inquiry-based approach, the problem-solving method is generally not a time-saving method.
• In fact, it often requires more time than traditional methods because students explore multiple possibilities, gather information, test solutions, and reflect on their findings.
• The focus is on deep understanding rather than quick coverage of content.
• While it is highly beneficial in promoting meaningful learning, it demands considerable classroom time for planning, execution, and discussion.

Hint

•  Helping in the development of scientific attitude is a key strength, as students learn to think logically, question assumptions, and make evidence-based conclusions.
• Training in the scientific method occurs naturally in this approach since learners go through steps such as problem identification, hypothesis formation, experimentation, and evaluation.
• Being a learner-centered method, it places the student at the core of the learning process, encouraging autonomy, curiosity, and active participation.

Hence, the correct answer is time saving method.

Explanation:

Given:

$P(A)=P(\frac{A}{B})=0.25$

and $P(\frac{B}{A})=0.5$

I. $P(\frac{B}{A})=\frac{P(A\cap B)}{P(A)}$

⇒ P(A∩B) = P(A) P(B|A)

⇒ P(A∩B) = 0.25 × 0.5 = 0.125

Now

⇒ $P(\frac{B}{A})=\frac{P(A\cap B)}{P(B)}$

⇒ $P(B)=\frac{P(A\cap B)}{P(\frac{A}{B})}$

⇒ $P(B)=\frac{0.125}{0.25}=0.5$

Now, P(A).P(B) = 0.25 × 0.5 = 0.125 = P(A∩B)

Thus  A and B are independent

II. $P(\bar{A}\cup \bar{B})=1–P(A\cap B)$

= 1 – 0.125 = 0.875

III. $P(\bar{A}\cap \bar{B})=1–P(A\cup B)$

= 1 – [P(A) + P(B) – P(A ∩ B)

=  1 – [0.25 + 0.5 – 0.125]

= = 1 – 0.625 = 0.375

So all statements I, II, and III are correct.

∴ Option (d) is correct.

The correct answer is Option 1.

Key Points

• Let the probability that A hits the target be P(A)=14" id="MathJax-Element-62-Frame" role="presentation" style="position: relative;" tabindex="0">P(A)=14$P(A)=\frac{1}{4}$ $P(A) = \frac{1}{4}$ .
• Let the probability that B hits the target be P(B)=25" id="MathJax-Element-63-Frame" role="presentation" style="position: relative;" tabindex="0">P(B)=25$P(B)=\frac{2}{5}$ $P(B) = \frac{2}{5}$ .
• The probability that neither A nor B hits the target is the product of their individual probabilities of missing the target.
• The probability that A misses the target is 1−P(A)=1−14=34" id="MathJax-Element-64-Frame" role="presentation" style="position: relative;" tabindex="0">1−P(A)=1−14=34$1-P(A)=1-\frac{1}{4}=\frac{3}{4}$ $1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$ .
• The probability that B misses the target is 1−P(B)=1−25=35" id="MathJax-Element-65-Frame" role="presentation" style="position: relative;" tabindex="0">1−P(B)=1−25=35$1-P(B)=1-\frac{2}{5}=\frac{3}{5}$ $1 - P(B) = 1 - \frac{2}{5} = \frac{3}{5}$ .
• The probability that neither hits the target is (34)×(35)=920" id="MathJax-Element-66-Frame" role="presentation" style="position: relative;" tabindex="0">(34)×(35)=920$\left(\frac{3}{4}\right)\times \left(\frac{3}{5}\right)=\frac{9}{20}$ $\left(\frac{3}{4}\right) \times \left(\frac{3}{5}\right) = \frac{9}{20}$ .
• The probability that at least one of them hits the target is 1−920=1120" id="MathJax-Element-67-Frame" role="presentation" style="position: relative;" tabindex="0">1−920=1120$1-\frac{9}{20}=\frac{11}{20}$ $1 - \frac{9}{20} = \frac{11}{20}$ .

Additional Information

• This problem involves the concept of complementary probability, which is useful when calculating the probability of at least one event occurring.In probability theory, the complement rule is used to determine the probability of an event not occurring.
• Option 1 is indeed correct as the final probability calculation matches the probability required.

Choice C is correct. The predicted increase in total fat, in grams, for every increase of 1 gram in total protein is represented by the slope of the line of best fit. Any two points on the line can be used to calculate the slope of the line as the change in total fat over the change in total protein. For instance, it can be estimated that the points (20,34) and (30,48) are on the line of best fit, and the slope of the line that passes through them is $\frac{48-34}{30-20}=\frac{14}{10}$, or 1.4. Of the choices given, 1.5 is the closest to the slope of the line of best fit.
Choices A, B, and D are incorrect and may be the result of incorrectly finding ordered pairs that lie on the line of best fit or of incorrectly calculating the slope.

Choice B is correct. From the shape of the cluster of points, the line of best fit should pass roughly through the points (1,200) and (2.5.350). Therefore, these two points can be used to find an approximate equation for the line of best fit. The slope of this line of best fit is therefore $\frac{y_2-y_1}{x_2-x_1}=\frac{350-200}{2.5-1}$, or 100. The equation for the line of best fit, in slope-intercept form, is y = 100 x + b for some value of b. Using the point (1,200), 1 can be substituted for x and 200 can be substituted for y: 200 = 100(1) + b, or b = 100. Substituting this value into the slope-intercept form of the equation gives y = 100 x + 100. Choice A is incorrect. The line defined by y = 200 x + 100 passes through the points (1,300) and (2,500), both of which are well above the cluster of points, so it cannot be a line of best fit. Choice C is incorrect. The line defined by y = 50 x + 100 passes through the points (1,150) and (2,200), both of which lie at the bottom of the cluster of points, so it cannot be a line of best fit. Choice D is incorrect and may result from correctly calculating the slope of a line of best fit but incorrectly assuming the y-intercept is at (0, 0).

Choice D is correct. The data in the scatterplot roughly fall in the shape of a downward-opening parabola; therefore, the coefficient for the x² term must be negative. Based on the location of the data points, the y- intercept of the parabola should be somewhere between 740 and 760. Therefore, of the equations given, the best model is y = -1.674x2 + 19.76x + 745.73.
Choices A and C are incorrect. The positive coefficient of the x2 term means that these equations each define upward-opening parabolas, whereas a parabola that fits the data in the scatterplot must open downward. Choice B is incorrect because it defines a parabola with a y-intercept that has a negative y-coordinate, whereas a parabola that fits the data in the scatterplot must have a y-intercept with a positive y-coordinate.

The correct answer is 285. The number of people predicted to visit the beach each day is represented by the y values of the line of best fit, and the average temperature, in degrees Celsius (°C), is represented by the x values. Since the slope of the line of best fit is approximately 57, the y-value, or the number of people predicted to visit the beach each day, increases by 57 for every x-value increase of 1, or every 1°C increase in average temperature. Therefore, an increase of in 5°C average temperature corresponds to a y-value increase of 57(5) = 285 additional people per day predicted to visit the beach. 

The correct answer is $\frac{9}{2}$. It's given that the scatterplot shows the relationship between the length of time y, in hours, a certain bird spent in flight and the number of days after January 11, x. Since January 13 is 2 days after January 11, it follows that January 13 corresponds to an x-value of 2 in the scatterplot. In the scatterplot, when x = 2, the corresponding value of y is 6. In other words, on January 13, the bird spent 6 hours in flight. Since January 15 is 4 days after January 11, it follows that January 15 corresponds to an x-value of 4 in the scatterplot. In the scatterplot, when x = 4, the corresponding value of y is 15. In other words, on January 15, the bird spent 15 hours in flight. Therefore, the average rate of change, in hours per day, of the length of time the bird spent in flight on January 13 to the length of time the bird spent in flight on January 15 is the difference in the length of time, in hours, the bird spent in flight divided by the difference in the number of days after January $11\text{, or }\frac{15-6}{4-2}$, which is equivalent to $\frac{9}{2}$. Note that 9/2 and 4.5 are examples of ways to enter a correct answer.

Choice D is correct. On the line of best fit, d increases from approximately 480 to 880 between t = 12 and t = 24. The slope of the line of best fit is the difference in d-values divided by the difference in t-values, which gives $\frac{880-480}{24-12}=\frac{400}{12}$, or approximately 33. Writing the equation of the line of best fit in slope-intercept form gives d = 33 t + b, where b is the y-coordinate of the y-intercept. This equation is satisfied by all points on the line, so d = 480 when t = 12. Thus, 480 = 33(12) + b, which is equivalent to 480 = 396 + b. Subtracting 396 from both sides of this equation gives b = 84. Therefore, an equation for the line of best fit could be d = 33 t + 84. 
Choice A is incorrect and may result from an error in calculating the slope and misidentifying the y coordinate of the y-intercept of the graph as the value of d at t = 10 rather than the value of d at t = 0. Choice B is incorrect and may result from using the smallest value of t on the graph as the slope and misidentifying the y-coordinate of the y-intercept of the graph as the value of d at rather than the value of d at t = 0. 
Choice C is incorrect and may result from misidentifying the y-coordinate of the y-intercept as the smallest value of d on the graph. 

Choice A is correct. The given equation is in slope-intercept form, or y = m x + b, where m is the value of the slope of the line of best fit. Therefore, the slope of the line of best fit is 0.096. From the definition of slope, it follows that an increase of 1 in the x-value corresponds to an increase of 0.096 in the y-value. Therefore, the line of best fit predicts that for each year between 1940 and 2010, the minimum wage will increase by 0.096 dollar per hour. 
Choice B is incorrect and may result from using the y-coordinate of the y-intercept as the average increase, instead of the slope. Choice C is incorrect and may result from using the 10-year increments given on the x axis to incorrectly interpret the slope of the line of best fit. Choice D is incorrect and may result from using the y-coordinate of the y-intercept as the average increase, instead of the slope, and from using the 10-year increments given on the x-axis to incorrectly interpret the slope of the line of best fit.