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# Learn UNIT VII: Statistics and probability with Free Lessons & Tips

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To find the probability of drawing a card with a number less than 10, we need to determine the total number of favorable outcomes (cards with numbers less than 10) and divide it by the total number of possible outcomes (total number of cards). read more

To find the probability of drawing a card with a number less than 10, we need to determine the total number of favorable outcomes (cards with numbers less than 10) and divide it by the total number of possible outcomes (total number of cards).

$Given:&space;-&space;Numbers&space;on&space;the&space;cards&space;range&space;from&space;6&space;to&space;15.&space;1.&space;**Total&space;number&space;of&space;possible&space;outcomes&space;(total&space;number&space;of&space;cards):**&space;Since&space;there&space;are&space;cards&space;marked&space;with&space;numbers&space;from&space;6&space;to&space;15,&space;the&space;total&space;number&space;of&space;cards&space;is&space;$$15&space;-&space;6&space;+&space;1&space;=&space;10$$.&space;2.&space;**Total&space;number&space;of&space;favorable&space;outcomes&space;(cards&space;with&space;numbers&space;less&space;than&space;10):**&space;Cards&space;marked&space;with&space;numbers&space;less&space;than&space;10&space;are:&space;6,&space;7,&space;8,&space;and&space;9.&space;There&space;are&space;$$10&space;-&space;6&space;=&space;4$$&space;such&space;cards.$

$Now,&space;let's&space;calculate&space;the&space;probability&space;of&space;drawing&space;a&space;card&space;with&space;a&space;number&space;less&space;than&space;10:&space;$&space;\text{Probability}&space;=&space;\frac{\text{Number&space;of&space;favorable&space;outcomes}}{\text{Total&space;number&space;of&space;possible&space;outcomes}}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{4}{10}&space;$&space;$&space;\text{Probability}&space;=&space;0.4&space;$&space;So,&space;the&space;probability&space;of&space;getting&space;a&space;card&space;with&space;a&space;number&space;less&space;than&space;10&space;is&space;$$0.4$$&space;or&space;$$40\%$$.$

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In a standard deck of 52 playing cards, there are 2 black kings: the king of spades and the king of clubs. Total number of possible outcomes (total number of cards): There are 52 cards in a standard deck. Total number of favorable outcomes (black kings): There are 2 black kings in the deck. read more

In a standard deck of 52 playing cards, there are 2 black kings: the king of spades and the king of clubs.

1. Total number of possible outcomes (total number of cards): There are 52 cards in a standard deck.

2. Total number of favorable outcomes (black kings): There are 2 black kings in the deck.
$Now,&space;let's&space;calculate&space;the&space;probability&space;of&space;drawing&space;a&space;black&space;king:&space;$&space;\text{Probability}&space;=&space;\frac{\text{Number&space;of&space;favorable&space;outcomes}}{\text{Total&space;number&space;of&space;possible&space;outcomes}}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{2}{52}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{1}{26}&space;$&space;So,&space;the&space;probability&space;of&space;drawing&space;a&space;black&space;king&space;from&space;a&space;well-shuffled&space;deck&space;of&space;52&space;cards&space;is&space;$$&space;\frac{1}{26}&space;$$.$

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To find the probability of getting a black ball, we need to consider the total number of balls in the bag and the number of black balls. read more

To find the probability of getting a black ball, we need to consider the total number of balls in the bag and the number of black balls.

$Given:&space;-&space;4&space;red&space;balls&space;-&space;6&space;black&space;balls&space;1.&space;**Total&space;number&space;of&space;balls:**&space;There&space;are&space;$$4&space;+&space;6&space;=&space;10$$&space;balls&space;in&space;total.&space;2.&space;**Number&space;of&space;black&space;balls:**&space;There&space;are&space;6&space;black&space;balls.&space;Now,&space;let's&space;calculate&space;the&space;probability&space;of&space;getting&space;a&space;black&space;ball:&space;$&space;\text{Probability}&space;=&space;\frac{\text{Number&space;of&space;black&space;balls}}{\text{Total&space;number&space;of&space;balls}}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{6}{10}&space;$&space;$&space;\text{Probability}&space;=&space;0.6&space;$&space;So,&space;the&space;probability&space;of&space;getting&space;a&space;black&space;ball&space;from&space;the&space;bag&space;is&space;$$0.6$$&space;or&space;$$60\%$$.$

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To find the probability of getting a number less than 3 when a die is thrown once, we need to determine the total number of favorable outcomes (rolling a number less than 3) and divide it by the total number of possible outcomes (rolling any number on the die). Given: The die has 6 faces numbered 1... read more

To find the probability of getting a number less than 3 when a die is thrown once, we need to determine the total number of favorable outcomes (rolling a number less than 3) and divide it by the total number of possible outcomes (rolling any number on the die).

Given:

• The die has 6 faces numbered 1 through 6.
1. Total number of possible outcomes: There are 6 possible outcomes when rolling a die.

2. Number of favorable outcomes (rolling a number less than 3): There are 2 favorable outcomes: rolling a 1 or rolling a 2.

Now, let's calculate the probability of getting a number less than 3:

$\fn_phv&space;$&space;\text{Probability}&space;=&space;\frac{\text{Number&space;of&space;favorable&space;outcomes}}{\text{Total&space;number&space;of&space;possible&space;outcomes}}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{2}{6}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{1}{3}&space;$&space;So,&space;the&space;probability&space;of&space;getting&space;a&space;number&space;less&space;than&space;3&space;when&space;a&space;die&space;is&space;thrown&space;once&space;is&space;$$&space;\frac{1}{3}&space;$$.$

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To find the probability of drawing a card with an even number, we need to determine the total number of even-numbered cards in the bag and divide it by the total number of cards in the bag. Given: Numbers on the cards range from 3 to 20. read more

To find the probability of drawing a card with an even number, we need to determine the total number of even-numbered cards in the bag and divide it by the total number of cards in the bag.

Given:

• Numbers on the cards range from 3 to 20.$\fn_phv&space;1.&space;**Total&space;number&space;of&space;possible&space;outcomes&space;(total&space;number&space;of&space;cards):**&space;Since&space;the&space;numbers&space;on&space;the&space;cards&space;range&space;from&space;3&space;to&space;20,&space;there&space;are&space;$$20&space;-&space;3&space;+&space;1&space;=&space;18$$&space;cards&space;in&space;total.&space;2.&space;**Number&space;of&space;favorable&space;outcomes&space;(even-numbered&space;cards):**&space;Even&space;numbers&space;between&space;3&space;and&space;20&space;are:&space;4,&space;6,&space;8,&space;...,&space;20.&space;There&space;are&space;9&space;even-numbered&space;cards&space;in&space;total.&space;Now,&space;let's&space;calculate&space;the&space;probability&space;of&space;drawing&space;a&space;card&space;with&space;an&space;even&space;number:&space;$&space;\text{Probability}&space;=&space;\frac{\text{Number&space;of&space;favorable&space;outcomes}}{\text{Total&space;number&space;of&space;possible&space;outcomes}}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{9}{18}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{1}{2}&space;$&space;So,&space;the&space;probability&space;that&space;the&space;number&space;on&space;the&space;card&space;taken&space;out&space;is&space;an&space;even&space;number&space;is&space;$$&space;\frac{1}{2}&space;$$.$
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To find the probability that two friends born in the year 2000 have the same birthday, we need to consider the total number of days in the year 2000 and the number of ways they could share a birthday. Given that they were born in the same year: Total number of days in the year 2000: Since the year... read more

To find the probability that two friends born in the year 2000 have the same birthday, we need to consider the total number of days in the year 2000 and the number of ways they could share a birthday.

Given that they were born in the same year:

1. Total number of days in the year 2000: Since the year 2000 was a leap year, it had 366 days.

2. Number of ways they could share a birthday: Each friend's birthday can fall on any of the 366 days in the year.
$Now,&space;let's&space;calculate&space;the&space;probability&space;that&space;they&space;have&space;the&space;same&space;birthday:&space;$&space;\text{Probability}&space;=&space;\frac{\text{Number&space;of&space;ways&space;they&space;could&space;share&space;a&space;birthday}}{\text{Total&space;number&space;of&space;days&space;in&space;the&space;year&space;2000}}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{366}{366}&space;$&space;$&space;\text{Probability}&space;=&space;1&space;$&space;So,&space;the&space;probability&space;that&space;two&space;friends&space;born&space;in&space;the&space;year&space;2000&space;have&space;the&space;same&space;birthday&space;is&space;$$1$$,&space;meaning&space;it&space;is&space;certain&space;that&space;they&space;have&space;the&space;same&space;birthday.$

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If each observation in a continuous frequency distribution is increased by 5, the median of the new data set will also increase by 5. This is because the median is the middle value of the data set, and adding the same constant to every observation simply shifts the entire data set by that constant. read more

If each observation in a continuous frequency distribution is increased by 5, the median of the new data set will also increase by 5. This is because the median is the middle value of the data set, and adding the same constant to every observation simply shifts the entire data set by that constant.$IGiven&space;that&space;the&space;original&space;median&space;is&space;21,&space;when&space;each&space;observation&space;is&space;increased&space;by&space;5,&space;the&space;new&space;median&space;will&space;be:&space;$&space;\text{New&space;median}&space;=&space;\text{Original&space;median}&space;+&space;5&space;$&space;$&space;\text{New&space;median}&space;=&space;21&space;+&space;5&space;$&space;$&space;\text{New&space;median}&space;=&space;26&space;$&space;So,&space;the&space;new&space;median&space;of&space;the&space;data&space;set&space;after&space;each&space;observation&space;is&space;increased&space;by&space;5&space;is&space;26.$

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Let's denote: b as the number of boys, g as the number of girls. Given: The average score of boys is 71. The average score of girls is 73. The average score of the school is 71.8. read more

Let's denote:

• b as the number of boys,
• g as the number of girls.

Given:

1. The average score of boys is 71.
2. The average score of girls is 73.
3. The average score of the school is 71.8.
$We&space;can&space;use&space;the&space;concept&space;of&space;weighted&space;averages&space;to&space;solve&space;this&space;problem.&space;The&space;total&space;score&space;of&space;all&space;the&space;students&space;is&space;the&space;sum&space;of&space;the&space;individual&space;scores&space;of&space;boys&space;and&space;girls&space;combined.&space;The&space;total&space;number&space;of&space;students&space;is&space;the&space;sum&space;of&space;the&space;number&space;of&space;boys&space;and&space;girls.&space;The&space;average&space;score&space;of&space;the&space;school&space;can&space;be&space;calculated&space;using&space;the&space;formula&space;for&space;weighted&space;averages:&space;$&space;\text{Average&space;score&space;of&space;the&space;school}&space;=&space;\frac{\text{Total&space;score&space;of&space;all&space;students}}{\text{Total&space;number&space;of&space;students}}&space;$&space;We&space;can&space;express&space;this&space;equation&space;using&space;the&space;given&space;information:&space;$&space;71.8&space;=&space;\frac{(71&space;\times&space;b)&space;+&space;(73&space;\times&space;g)}{b&space;+&space;g}&space;$$
$Given&space;that&space;the&space;number&space;of&space;boys&space;plus&space;the&space;number&space;of&space;girls&space;is&space;the&space;total&space;number&space;of&space;students,&space;we&space;have:&space;$&space;b&space;+&space;g&space;=&space;\text{Total&space;number&space;of&space;students}&space;$&space;We&space;know&space;that&space;the&space;average&space;score&space;of&space;boys&space;is&space;71&space;and&space;the&space;average&space;score&space;of&space;girls&space;is&space;73,&space;so:&space;$&space;71&space;=&space;\frac{(71&space;\times&space;b)}{b&space;+&space;g}&space;$&space;$&space;73&space;=&space;\frac{(73&space;\times&space;g)}{b&space;+&space;g}&space;$&space;Now,&space;let's&space;solve&space;this&space;system&space;of&space;equations&space;to&space;find&space;the&space;values&space;of&space;$$b$$&space;and&space;$$g$$,&space;and&space;then&space;determine&space;the&space;ratio&space;$$b:g$$.$
4. $Given&space;that&space;the&space;number&space;of&space;boys&space;plus&space;the&space;number&space;of&space;girls&space;is&space;the&space;total&space;number&space;of&space;students,&space;we&space;have:&space;$&space;b&space;+&space;g&space;=&space;\text{Total&space;number&space;of&space;students}&space;$&space;We&space;know&space;that&space;the&space;average&space;score&space;of&space;boys&space;is&space;71&space;and&space;the&space;average&space;score&space;of&space;girls&space;is&space;73,&space;so:&space;$&space;71&space;=&space;\frac{(71&space;\times&space;b)}{b&space;+&space;g}&space;$&space;$&space;73&space;=&space;\frac{(73&space;\times&space;g)}{b&space;+&space;g}&space;$&space;Now,&space;let's&space;solve&space;this&space;system&space;of&space;equations&space;to&space;find&space;the&space;values&space;of&space;$$b$$&space;and&space;$$g$$,&space;and&space;then&space;determine&space;the&space;ratio&space;$$b:g$$.&space;First,&space;we&space;can&space;find&space;the&space;total&space;number&space;of&space;students&space;($$b&space;+&space;g$$)&space;from&space;the&space;given&space;average&space;score&space;of&space;the&space;school:&space;$&space;71.8&space;=&space;\frac{(71&space;\times&space;b)&space;+&space;(73&space;\times&space;g)}{b&space;+&space;g}&space;$&space;$&space;71.8&space;\times&space;(b&space;+&space;g)&space;=&space;(71&space;\times&space;b)&space;+&space;(73&space;\times&space;g)&space;$&space;$&space;71.8b&space;+&space;71.8g&space;=&space;71b&space;+&space;73g&space;$&space;$&space;71.8g&space;-&space;73g&space;=&space;71b&space;-&space;71.8b&space;$&space;$&space;-1.2g&space;=&space;-0.8b&space;$&space;$&space;3g&space;=&space;2b&space;$&space;From&space;this,&space;we&space;get&space;$$3g&space;=&space;2b$$,&space;or&space;equivalently,&space;$$b&space;=&space;\frac{3}{2}g$$.&space;We&space;know&space;that&space;the&space;ratio&space;$$b:g$$&space;is&space;$$3:2$$.&space;So,&space;the&space;ratio&space;of&space;the&space;number&space;of&space;boys&space;to&space;the&space;number&space;of&space;girls&space;who&space;appeared&space;in&space;the&space;examination&space;is&space;$$3:2$$.$
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To find the correct mean, we first need to adjust the sum of the observations by adding the difference between the correct value of 125 and the wrongly noted value of 25. Then we can divide this adjusted sum by the total number of observations. read more

To find the correct mean, we first need to adjust the sum of the observations by adding the difference between the correct value of 125 and the wrongly noted value of 25. Then we can divide this adjusted sum by the total number of observations.

$Given:&space;-&space;The&space;mean&space;calculated&space;using&space;the&space;wrong&space;value&space;was&space;60.&space;-&space;The&space;wrong&space;value&space;noted&space;was&space;25&space;instead&space;of&space;125.&space;Let's&space;denote:&space;-&space;$$x$$&space;as&space;the&space;sum&space;of&space;the&space;correct&space;values&space;of&space;all&space;observations,&space;-&space;$$n$$&space;as&space;the&space;total&space;number&space;of&space;observations.&space;The&space;sum&space;of&space;the&space;observations&space;with&space;the&space;wrong&space;value&space;noted&space;was&space;$$x&space;-&space;125&space;+&space;25&space;=&space;x&space;-&space;100$$.&space;The&space;mean&space;calculated&space;using&space;the&space;wrong&space;value&space;was&space;60,&space;so:&space;$&space;\text{Mean}&space;=&space;\frac{x&space;-&space;100}{n}&space;=&space;60&space;$$

$Now,&space;we&space;can&space;solve&space;for&space;$$x$$&space;using&space;this&space;equation:&space;$&space;x&space;-&space;100&space;=&space;60n&space;$&space;$&space;x&space;=&space;60n&space;+&space;100&space;$&space;Now,&space;we&space;need&space;to&space;find&space;the&space;correct&space;mean&space;using&space;the&space;correct&space;sum&space;of&space;observations&space;$$x$$.&space;The&space;correct&space;mean&space;($$M$$)&space;can&space;be&space;calculated&space;as:&space;$&space;M&space;=&space;\frac{x}{n}&space;=&space;\frac{60n&space;+&space;100}{n}&space;$&space;$&space;M&space;=&space;60&space;+&space;\frac{100}{n}&space;$&space;So,&space;the&space;correct&space;mean&space;is&space;$$60&space;+&space;\frac{100}{n}$$.&space;We&space;need&space;the&space;total&space;number&space;of&space;observations&space;($$n$$)&space;to&space;calculate&space;the&space;correct&space;mean.&space;If&space;you&space;have&space;the&space;value&space;of&space;$$n$$,&space;please&space;provide&space;it&space;so&space;that&space;we&space;can&space;find&space;the&space;correct&space;mean.$

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To show that the mode of the combined series obtained by combining the two series S1 and S2 is different from that of S1 and S2 taken separately, we first need to find the mode of each series individually and then compare it with the mode of the combined series. read more

To show that the mode of the combined series obtained by combining the two series S1 and S2 is different from that of S1 and S2 taken separately, we first need to find the mode of each series individually and then compare it with the mode of the combined series.

$Let's&space;start&space;by&space;finding&space;the&space;mode&space;of&space;$$S1$$&space;and&space;$$S2$$&space;separately:&space;For&space;$$S1$$:&space;-&space;The&space;mode&space;of&space;a&space;set&space;of&space;data&space;is&space;the&space;number&space;that&space;appears&space;most&space;frequently.&space;-&space;In&space;$$S1$$,&space;the&space;number&space;9&space;appears&space;three&space;times,&space;which&space;is&space;more&space;frequent&space;than&space;any&space;other&space;number.&space;-&space;So,&space;the&space;mode&space;of&space;$$S1$$&space;is&space;9.&space;For&space;$$S2$$:&space;-&space;In&space;$$S2$$,&space;the&space;numbers&space;7&space;and&space;8&space;both&space;appear&space;three&space;times,&space;which&space;is&space;more&space;frequent&space;than&space;any&space;other&space;number.&space;-&space;So,&space;$$S2$$&space;has&space;two&space;modes:&space;7&space;and&space;8.$

$Now,&space;let's&space;combine&space;$$S1$$&space;and&space;$$S2$$&space;into&space;one&space;series:&space;Combined&space;series:&space;$$S1&space;\cup&space;S2$$&space;$&space;S1&space;\cup&space;S2&space;=&space;\{3,&space;5,&space;8,&space;8,&space;9,&space;12,&space;13,&space;9,&space;9,&space;7,&space;4,&space;7,&space;8,&space;7,&space;8,&space;13\}&space;$&space;To&space;find&space;the&space;mode&space;of&space;the&space;combined&space;series,&space;we&space;need&space;to&space;determine&space;which&space;number&space;appears&space;most&space;frequently.&space;-&space;In&space;the&space;combined&space;series,&space;the&space;number&space;9&space;appears&space;three&space;times,&space;and&space;the&space;numbers&space;7&space;and&space;8&space;each&space;appear&space;four&space;times.&space;-&space;So,&space;the&space;mode&space;of&space;the&space;combined&space;series&space;is&space;either&space;7&space;or&space;8.&space;Therefore,&space;we&space;can&space;see&space;that&space;the&space;mode&space;of&space;the&space;combined&space;series&space;($$S1&space;\cup&space;S2$$)&space;is&space;different&space;from&space;that&space;of&space;$$S1$$&space;and&space;$$S2$$&space;taken&space;separately.&space;In&space;$$S1$$,&space;the&space;mode&space;is&space;9.&space;In&space;$$S2$$,&space;the&space;modes&space;are&space;7&space;and&space;8.&space;In&space;$$S1&space;\cup&space;S2$$,&space;the&space;mode&space;is&space;7&space;or&space;8.$

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