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#### $80$ bulbs are selected at random from a lot and their life time (in hrs) is recorded in the form of a frequency table given below : Life time (in hours) 300 500 700 900 1100 Frequency 10 12 23 25 10 The probability that bulbs selected randomly from the lot has life less than $900$ hours is : $(A)\ \frac{11}{40}$ $(B)\ \frac{5}{16}$ $(C)\ \frac{7}{16}$$(D) \ \frac{9}{16}$

$\text {P}=\frac{\text {favourable outcome}}{\text {Total outcomes}}$

Bulb’s life is given as less than 900 hours = Bulbs with life 300, 500 and 700 hours

From the table, we can see that bulbs with life less than 900 hours =10+12+23

$\text {P}=\frac{\text {favourable outcome}}{\text {Total favourable outcomes}}=\frac{45}{10+12+23+25+10}$

$=\frac{45}{80}=\frac{9}{16}$

Therefore option (D) is correct.

#### 80 bulbs are selected at random from a lot and their life time (in hrs) is recorded in the form of a frequency table given below : Life time (in hours) 300 500 700 900 1100 Frequency 10 12 23 25 10 One bulb is selected at random from the lot. The probability that its life is 1150 hours, is $(A)\ \frac{1}{80}$ $(B)\ \frac{7}{16}$ $(c)\ 0$ $(D)\ 1$

$\text {P}=\frac{\text {favourable outcome}}{\text {Total outcomes}}$

Bulb’s life is given as $1150$ hours.

From the table, we can see that no bulb has life as$1150$ hours.

So favourable outcomes $=0$

Required probability $=\frac{0}{10+12+23+25+10}=0$

Therefore option (C) is correct.

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#### Two coins are tossed 1000 times and the outcomes are recorded as below : Number of heads 2 1 0 Frequency 200 550 250 Based on this information, the probability for at most one head is: $(A)\ \frac{1}{5}$ $(B)\ \frac{1}{4}$ $(C)\ \frac{4}{5}$ $(D)\ \frac{3}{4}$

Total number of outcomes =200+550+250=1000

Number of favourable outcomes =250+550=800

Probability will be

$=\frac{800}{1000}=\frac{4}{5}$

Therefore option (C) is correct.

#### In a medical examination of students of a class, the following blood groups are recorded: Blood Group A AB B O Number of students 10 13 12 5 A student is selected at random from the class. The probability that he/she has blood group B, is: $(A)\ \frac{1}{4}$ $(B)\ \frac{13}{40}$$(C)\ \frac{3}{10}$$(D)\ \frac{1}{8}$

$\text {P}=\frac{\text {favourable outcomes}}{\text {Total possible outcome}}$

Number of students with blood group B =12

Total number of students =10+13+12+5

Required probability

$=\frac{12}{10+13+12+5}=\frac{12}{40}=\frac{3}{10}$

Therefore option (B) is correct.

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#### In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he/she does not like to eat potato chips is:$\\(A) 0.25\\ (B) 0.50\\ (C) 0.75\\ (D) 0.80$

In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips.

If a child is selected at random, the probability that he/she likes to eat potato chips is

$=\frac{91}{364}$

If a child is selected at random, the probability that he/she does not like to eat potato chips is

$=1-\frac{91}{364}$

$=\frac{364-91}{361}=\frac{273}{361}=0.756$

In the options given, the closest to 0.756 is 0.75

Therefore option (C) is correct.

#### In a sample study of 642 people, it was found that 514 people have a high school certificate. If a person is selected at random, the probability that the person has a high school certificate is :$\\(A) 0.5\\ (B) 0.6\\ (C) 0.7\\ (D) 0.8$

We will use the formula of probability:-

$\text {P}=\frac{\text {favourable outcome}}{\text {Total possible outcome}}$

Total possible outcomes =642

Favourable outcomes =514

Required probability

$=\frac{514}{642}=0.800$

So if a person is selected at random, the probability that the person has a high school certificate is 0.8

Therefore option (D) is correct.

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#### Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is(A) 14(B) 15(C) 16(D) 17

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

Given data, 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15

Arranging in ascending order

14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20

Here, 15 occurs 5 times, which is the most number of times in the given set of values.

Mode =15

Therefore option (B) is correct.

#### Median of the following numbers : 4, 4, 5, 7, 6, 7, 7, 12, 3 is$\\(A) 4\\ (B) 5\\ (C) 6\\ (D) 7$

Given numbers: 4, 4, 5, 7, 6, 7, 7, 12, 3

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

Firstly, we will arrange the numbers in ascending order –

$3, 4, 4, 5, 6, 7, 7, 7, 12$

There are $9$ number of observations.

For odd number of observations, middle term

$=\left ( \frac{n+1}{2} \right )^{th}\text {term}=\left ( \frac{9+1}{2} \right )^{th}\text {term}=\left ( \frac{10}{2} \right )^{th}\text {term}=5^{th}\text {term}$

5th term is 6, hence it is the required median.

Therefore option (C) is correct.

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#### For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are respectively :(A) upper limits of the classes(B) lower limits of the classes(C) class marks of the classes(D) upper limits of preceding classes

For drawing a frequency polygon of a continuous frequency distribution, we plot the frequency of the classes on the ordinates and the class marks of the classes on the abscissae.

Class mark is the mid value or the central value of a class

It is calculated as follows:

$\frac{\text {upper limit+lower limit}}{2}$

For example

 Class interval Class marks (x-coordinate) Frequency (y-coordinate) 10-20 15 7 20-30 25 8 30-40 35 9

Therefore option (C) is correct.

#### The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is$\\(A) 45\\ (B) 49.5\\ (C) 54\\ (D) 56$

Given terms : 78, 56, 22, 34, 45, 54, 39, 68, 54, 84

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

In ascending order:

$22, 34, 39, 45, 54, 54, 56, 68, 78, 84$

Number of terms

$=10\left ( \text {even} \right )$

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

$\left ( \frac{n}{2} \right )^{th}\text {term}=\left ( \frac{10}{2} \right )^{th}=5^{th}\text {term}=54$

$\text {Median}=\frac{\text {term}+\text {6th term}}{2}=\frac{108}{2}=54$

Therefore option (C) is correct.