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A recent survey found that the ages of workers in a factory is distributed as follows: Age (in years) 20-29 30-39 40-49 50-59 60 and above Number of workers 38 27 86 46 3 If a person is selected at random, find the probability that the person is:(i) 40 years or more(ii) under 40 years(iii) having age from 30 to 39 years(iv) under 60 but over 39 years

$\text{Probability}=\frac{\text {Favourable outcomes}}{\text {Total number of events}}$

Here, total events = total number of workers= 38 + 27 + 86 + 46 + 3 = 200

(i) p (person is 40 years or more) = p(person having age 40 to 49 years) + p (person having age 50 to 59 years) + p (having age 60 and above)

$=\frac{86}{200}+\frac{46}{200}+\frac{3}{200}=\frac{135}{200}=0.675$

(ii) p(person is under 40 years) = p(person having age 20 to 29 years) + p(person having age 30 to 39 years)

$=\frac{38}{200}+\frac{27}{200}=\frac{65}{200}=0.325$

Hence the different age group decided the work.

$\text{(iii) p(having age from 30 to 39 years) }=\frac{27}{200}=0.135$

(iv) p(under 60 but over 39 years) = p(person having age 40 to 49 years) + p (person having age 50 to 59 years)

$=\frac{86}{200}+\frac{46}{200}=\frac{132}{200}=0.66$

Over the past 200 working days, the number of defective parts produced by a machine is given in the following table: Number of defective parts 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Days 50 32 22 18 12 12 10 10 10 8 6 6 2 2 Determine the probability that tomorrow’s output will have(i) no defective part(ii) at least one defective part(iii) not more than 5 defective parts(iv) more than 13 defective parts

Probability is defined as $=\frac{\text {Favourable outcomes}}{\text {Total number of events}}$

Here, total events = total number of working days $= 200$

(i) no defective part

Favourable outcomes $= 50$ days $= 50$

p (no defective part)$=\frac{50}{200}=0.25$

(ii) Probability that at least one defective part = 1 - probability that no defective part

p (no defective part)$=\frac{50}{200}=0.25$

p (at least one defective part) $=1-\frac{50}{200}=0.75$

(iii) not more than 5 defective parts = 0 or 1 or 2 or 3 or 4 or 5 defective parts

p(not more than 5 defective parts) = p(no. defective part) + p(1 defective part) + p(2 defective part) + p(3 defective part) + p(4 defective part) + p(5 defective part)

$=\frac{50}{200}+\frac{32}{200}+\frac{22}{200}+\frac{18}{200}+\frac{12}{200}+\frac{12}{200}$

$=\frac{50+32+22+18+12+12}{200}=\frac{146}{200}$

$=\frac{73}{200}=0.73$

(iv) more than 13 defective parts = not possible

Favourable outcomes = 0

p (more than 13 defective parts) = 0

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Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table: Number of defective bulbs 0 1 2 3 4 5 6 more than 6 Frequency 400 180 48 41 18 8 3 2 One carton was selected at random. What is the probability that it has(i) no defective bulb?(ii) defective bulbs from 2 to 6?(iii) defective bulbs less than 4?

Probability is defined as $=\frac{\text {Favourable outcomes}}{\text {Total number of events}}$

Here, total events = total cartons $= 700$

(i) no defective bulb

Favourable outcomes $= 400$

p (cartoon has no defective bulb) =$\frac{400}{700}=\frac{4}{7}$

(ii) defective bulbs from 2 to 6 = 2 or 3 or 4 or 5 or 6 defective bulbs

Favourable outcomes $= 48 + 41 + 18 + 8 + 3 = 118$

p(defective bulb from 2 to 6) $=\frac{118}{700}=\frac{59}{350}$

(iii) defective bulbs less than 4 = defective bulbs equal to 0 or 1 or 2 or 3

Favourable outcomes $= 400 + 180 + 48 + 41 = 669$

p(defective bulbs less than 4)$=\frac{669}{700}$

Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table: Sum Frequency 2 14 3 30 4 42 5 55 6 72 7 75 8 70 9 53 10 46 11 28 12 15 If the dice are thrown once more, what is the probability of getting a sum(i) 3?                                                    (ii) more than 10?(iii) less than or equal to 5?                (iv) between 8 and 12?

$\text{Probability }=\frac{\text {Favourable outcomes}}{\text {Total number of events}}$

Here, total events  = 14 + 30 + 42 + 55 + 72 + 75 + 70 + 53 + 46 + 28 +15 = 500

(i) probability of getting a sum = 3

Favourable events = 30

$\text{P[of getting sum 3]}=\frac{30}{500}=0.06$

(ii) probability of getting a sum more than 10

$\text{Favourable events}= 28 + 15 = 43$

$\text{P[of getting sum 10]}=\frac{43}{500}=0.086$

(iii) probability of getting a sum less than or equal to 5

Favourable events = 14 + 30 + 42 + 55

$\text{p[sum less than or equal to 5]}=\frac{141}{500}=\frac{28.2}{100}=2.282$

(iv) probability of getting a sum between 8 and 12

Favourable events = 53 + 46 + 28 = 127

$\text{p[sum between 8 and 12]}=\frac{127}{500}=0.254$

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A company selected 4000 households at random and surveyed them to find out a relationship between income level and the number of television sets in a home. The information so obtained is listed in the following table: Monthly income (in Rs) Number of Televisions/household   0 1 2 Above 2 <10000 20 80 10 0 10000-14999 10 240 60 0 15000-19999 0 380 120 30 20000-24999 0 520 370 80 25000 and above 0 1100 760 220 Find the probability:(i) of a household earning Rs 10000 – Rs 14999 per year and having exactly one television.(ii) of a household earning Rs 25000 and more per year and owning 2 televisions.(iii) of a household not having any television.

Here, total events =4000

(i) A household earning Rs 10000 – Rs 14999 per year and having exactly one television.

Favourable outcomes =240

$\text{Required probability}=\frac{240}{4000}=\frac{3}{50}$

(ii) A household earning Rs 25000 and more per year and owning 2 televisions.

Favourable outcomes =760

$\text{Required probability}=\frac{760}{4000}=0.19$

(iii) A household not having any television.

Favourable outcomes =20+10

$\text{Required probability}=\frac{30}{4000}=\frac{3}{400}$

In Fig. 14.2, there is a histogram depicting the daily wages of workers in a factory. Construct the frequency distribution table.

Based on the given bar graph :

 Class-Interval Frequency 150-200 50 200-250 30 250-300 35 300-350 20 350-400 10 Total workers 145

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The points scored by a basketball team in a series of matches are as follows:$17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28$Find the median and mode for the data.

Answer : [median = 12 and mode = 10]

Solution. To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.

Here total elements, n = 16 (even)

The terms as arranged in ascending order:

$2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 48$

Number of observation = 16 (even number)

Now using formula of median in case number of terms is even

$\Rightarrow$ Median $=\frac{\left ( \frac{n}{2} \right )^{th}\text {obs}+\left ( \frac{n}{2}+1 \right )^{th}\text {obs}}{2}$

$=\frac{\left ( \frac{16}{2} \right )^{th}\text {obs}+\left ( \frac{16}{2}+1 \right )^{th}\text {obs}}{2}$

So median $=\frac{\left ( 8^{th}\text {obs.}+9^{th}obs. \right )}{2}$

$\Rightarrow \frac{10+14}{2}=\frac{24}{2}$

Median $=12$

Mode is the value that occurs the most number of times in a given set of values

Now mode is 10 because it is the most repeating number.

Hence, median = 12 and mode = 10.

Ten observations 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 are written in an ascending order. The median of the data is 24. Find the value of x.

Here, the observations are :

$6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43$

To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.

The terms are already given in ascending order so we have to find the middle term.

Number of terms, n=10(even)

$\text{So, Median = average of}\left ( \frac{n}{2} \right )^{th}$ $\text{ and}\left ( \frac{n}{2} +1\right )^{th}\text{term}$

$\Rightarrow$ Median =24            (given)

$\text{Median}=\frac{\left ( \frac{n}{2} \right )^{th}\text {observation}+\left ( \frac{n}{2}+1 \right )^{th}\text {observation}}{2}$

$24=\frac{\left ( \frac{10}{2} \right )^{th}\text {obs.}+\left ( \frac{10}{2}+1 \right )\text {obs.}}{2}$

$24=\frac{5^{th}\text {obs.}+6^{th}\text {obs.}}{2}$

$24=\frac{x+1+2x-13}{2}$

$\Rightarrow 2 \times 24=x+1+2x-13$

$\Rightarrow 48=3x-12$

$\Rightarrow 48+12=3x$

$\Rightarrow 60=3x$

$\Rightarrow x=20$

Hence, the correct answer is 20.

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Mean of 50 observations was found to be 80.4. But later on, it was discovered that 96 was misread as 69 at one place. Find the correct mean.

$\left [ 80.94 \right ]$

$\text{Mean}=\frac{\text {sum of all observations}}{\text {No. of observations}}$

Mean of 50 observations was found to be 80.4 (Incorrect mean)

The incorrect sum of all the numbers = Incorrect mean $\times$ Total numbers

$=80.4 \times 50=4020$

It was discovered that 96 was misread as 69 at one place

Hence the correct sum of all the numbers

$= 4020 - 69 + 96 = 4047$

So correct mean

$\frac{4047}{50}=80.94$

Hence, the correct mean is 80.94

A class consists of 50 students out of which 30 are girls. The mean of marks scored by girls in a test is 73 (out of 100) and that of boys is 71. Determine the mean score of the whole class.

A class consists of 50 students out of which 30 are girls.

The mean marks of 30 girls =73

Total score of 30 girls

$= 73 \times 30 = 2190.$

Now, number of boys are

$50 - 30 = 20.$

The mean marks of 20 boys =71

Total score of boys

$= 71 \times 20 = 1420$

$\Rightarrow$Here, the total score of whole class = Total score of all girls + Total score of all boys

$= 2190 + 1420 = 3610.$

$\text{Mean of all students}=\frac{\text{Total marks of all students} }{\text {Total number of students}}$

$=\frac{3610}{50}=72.2$

Hence, the mean of whole class is 72.2.