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Become a Probability & Statistics Master

 

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What you’ll learn

  • Visualizing data
  • Analyzing data

  • Data distributions

  • Probability
  • Discrete random variables
  • Sampling
  • Hypothesis testing

Requirements

  • Fundamentals of math
  • Basic algebra

Description

HOW BECOME A PROBABILITY & STATISTICS MASTER IS SET UP TO MAKE COMPLICATED MATH EASY

This 125-lesson course includes video explanations of everything from Probability & Statistics, and it includes more than 35 quizzes (with solutions!) to help you test your understanding along the way. Become a Probability & Statistics Master is organized into the following sections:

  • Visualizing data
  • Analyzing data
  • Data distributions
  • Probability
  • Discrete random variables
  • Sampling
  • Hypothesis testing

HERE’S WHAT SOME STUDENTS HAVE TOLD ME:

“Krista is an experienced teacher who offers Udemy students complete subject matter coverage and efficient and effective lessons/learning experiences. She not only understands the course material, but also selects/uses excellent application examples for her students and presents them clearly and skillfully using visual teaching aids/tools.” – John

“Really good, thorough, well explained lessons.” – Scott F.

“This is my second course (algebra previously) from Ms. King’s offerings. I enjoyed this course and learned a lot! Each video explains a concept, followed by the working of several examples. I learned the most by listening to Ms King’s teaching of the concept, stopping the video, and then attempting to work the example problems. After working the problems, then watching her complete the examples, I found that I really retained the concepts. A great instructor!” – Charles M.

Who is the target audience?

  • Current probability and statistics students, or students about to start probability and statistics who are looking to get ahead
  • Homeschool parents looking for extra support with probability and statistics
  • Anyone who wants to study math for fun after being away from school for a while

96 lectures

08:44:42

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A. about 60% of florida residents believe that florida is a nice place to live. suppose that six randomly selected florida residents are interviewed. what is the probability that at least one resident does not think that florida is a nice place to live?

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At least 6, means: either 1 and 2 and 3 and 4 and 5 and 6

Probability(that Florida is a nice place to live) = 60% = 0.6
If  at least  6 randomly selected like Florida:
P(at lest 6 LIKE Florida) = (0.6)⁶ = 0.04665
P(at least 6 DON’T LIKE Florida) = 1-0.04665 = 0.9533

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A family has 8 girls and 4 boys. A total of 2 children must be chosen to speak on the behalf of the family at a local benefit. What is the probability that 1 girl and 1 boy will be chosen?

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A family has 8 girls and 4 boys. A total of 2 children must be chosen to speak on the behalf of the family at a local benefit. What is the probability that 1 girl and 1 boy will be chosen?

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Assume there are 365 days in a year 1.) What is the probability that ten students in a class have different birthdays? 2.) What is the probability that at least ten students in a class share their birthday

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Assume there are 365 days in a year 1.) What is the probability that ten students in a class have different birthdays? 2.) What is the probability that at least ten students in a class share their birthday

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19. You roll two standard number cubes. What is the probability that the sum is odd, given than one of the number cubes shows a 1? Show your work.

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there are 6 numbers per cube,

 the chance of rolling one and getting the number 1 is a 1/6 probability

for the sum to be odd, you need to roll either a 2, 4 or 6 on the second cube so that gives you a 3/6 probability

 so 1/6 x 3/6 = 3/36 which can be reduced to 1/12

 so you have a 1/12 probability

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bridge hand is made up of 13 cards from a deck of 52. Find the probability that a hand chosen at random contains at least 1 nine. The probability that a bridge hand chosen at random contains at least 1 nine is:

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C(n,r) = n!/(r!(n-r)!)

You take the number of possible 13 card hands with no 9 in them and subtract that from the total number of possible 13 card hands 9’s included.

so C(52,13) – C(48,13)
The number of all possible 13 card hands is:
52!/13!(52-13)! or 52!/13!*39! which is 635,013,559,600

The number of all possible 13 card hands with no 9s is:

48!/13!(48-13)!  or 48!/13!*35! = 192,928,249,296

The difference is 635,013,559,600 – 192,928,249,296 = 442,085,310,304

So 442,085,310,304 out of 635,013,559,600 hands will have at least 1 nine.  The ratio of  442,085,310,304 to 635,013,559,600 is 0.696182472…  So roughly 70% of the time 

27,630,331,894/39,688,347,475 is the best I could do for a whole number ratio.

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The 4 aces are removed from a deck of cards. A coin is tossed and one of the aces is chosen. What is the probability of getting heads on the coin and the ace of hearts?

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The 4 aces are removed from a deck of cards. A coin is tossed and one of the aces is chosen. What is the probability of getting heads on the coin and the ace of hearts?

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Brenda throws a dart at this square-shaped target: Part A: Is the probability of hitting the black circle inside the target closer to 0 or 1? Explain your answer. (5 points) Part B: Is the probability of hitting the white portion of the target closer to 0 or 1? Explain your answer. (5 points)

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Either way.  The probability of hitting the circle is:

P(C)=Area of circle divided by area of square

P(W)=(area of square minus area of circle divided by area of square

P(C)=(πr^2)/s^2

P(W)=(s^2-πr^2)/s^2

Okay with know dimensions, r=1 (because r=d/2 and d=2 so r=1), s=11 we have:

P(inside circle)=π/121  (≈0.0259  or 2.6%)

P(outside circel)=(121-π)/121  (≈0.9744 or 97.4%)

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Three cards are chosen from a standard deck of 52 playing cards with replacement. What is the probability every card will be an Ace? Can someone please help

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Three cards are chosen from a standard deck of 52 playing cards with replacement. What is the probability every card will be an Ace? Can someone please help

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Consider a political discussion group consisting of 9 ​Democrats, 7 ​Republicans, and 3 Independents. Suppose that two group members are randomly​ selected, in​ succession, to attend a political convention. Find the probability of selecting no Democrats

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Well, first you would add up how many objects you have to select from

9 democrats + 7 Republicans + 3 Independents = 19 total people
so then you would put how many people you can have over that number
so since we don’t want any Democrats it would be 10/19 (.52)
and that tells you the chances of not picking one the first time

then, since that person cannot be selected again, we remove them from the pool, making it only 18 people to pick from and 9 people you want picked so once again you put 9/18 (.50), then you take those, (in decimal form) and multiply them (.26) then convert it to a fraction by moving the decimal to the right two and have 26% chance of no Democrats 

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The body temperatures of adults are normally distributed with a mean of 98.6° f and a standard deviation of 0.60° f. if 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° f. 0.0228 0.8188 0.9360 0.9772

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Here the population standard deviation is 0.60 degree F.  If a sample of 36 adults is randomly selected, that results in a sample standard deviation of 0.60 degree F divided by the square root of 36:  0.10 degree F.

The probability in question is the area under the standard normal probability distribution between 98.4 degree F and infinity, and intuitively you can detect that this will be more than 0.5 (corresponding to 50%).

Convert 98.4 degrees F to a z-score, using the sample standard deviation (0.10 degree F).  That z score is 
       98.4-98.6
z = ————–   =  -0.20/0.10 = -2
           0.10

We need to determine the area under the standard normal curve to the right of z=-2.  Use a table of z-scores to do this, or use your calculator’s built-in probability functions.  My result is 98.21% (corresponding to an area of 0.9821).

With my calculator I can find this probability using the following command:

normalcdf(-2,100000,0.10).

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When a standard pair of six sided dice are rolled, what is the probability of getting a 13?

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When a standard pair of six sided dice are rolled, what is the probability of getting a 13?

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If you have a bag that contains 24 marbles: 4 red, 12 blue, 6 black, 2 green. what is the probability of randomly picking one marble that is black?

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Parallel lines…means ur equations will have the same slope and different y int’s.

x – 2y = 8….-2y = -x + 8….y = 1/2x – 4…slope is 1/2, y int is -4
2x + 4y = 12…4y = -2x + 12….y = -1/2x + 3…slope is – 1/2, y int is 3
not this one…different slopes

x – 2y = 8…slope is 1/2, y int is -4
2x – 4y = 12…-4y = -2x + 12…y = 1/2x – 3….slope is 1/2, y int is -3
same slope, different y int’s…..these are parallel lines..but this is not ur graph because the graph has a negative slope.

x + 2y = 8….2y = -x + 8….y = -1/2x + 4..slope is -1/2, y int is 4
2x + 4y = 12…4y = -2x + 12….y = -1/2x + 3…slope is -1/2, y int is 3
same slope, different y int’s…parallel lines…and ur y int’s match the graph…this is ur answer <===

x + 2y = 8…slope is -1/2, y int is 4
2x – 4y = 12..slope is 1/2, y int is 3
different slopes, different y int….this has 1 solution and ur lines are not parallel…not this one

therefore, ur answer is : 3rd answer choice

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You are dealt one card from a standard 52-card deck. find the probability of being dealt an ace or a 9

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Answer:

(C)0.20

Step-by-step explanation:

We are given that we have to make five-letter codes that use the letters A, F, E, R, and M without repeating any letter.

Thus, total number of letters=5

Now, the probability that a randomly chosen code starts with A is given as:

P=frac{Favourable outcomes}{Total number of outcomes}

P=frac{1}{5}

P=0.20

Therefore, the the probability that a randomly chosen code starts with A is 0.20.

Hence, option C is correct.

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A bag contains 10 marbles: 4 are green, 4 are red, and 2 are blue. Heather chooses a marble at random, and without putting it back, chooses another one at random. What is the probability that both marbles she chooses are blue? Write your answer as a fraction in simplest form.

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Employee                                 Mary      Zoe         Greg         Ann           Tom

Cumulative Pay                       $6,800   $10,500  $8,400    $66,000   $4,700

Pay subject to FICA S.S.         $421.60  $651.00  $520.80 $4092.00 $291.40
6.2%, (First $118,000)

Pay subject to FICA Medicare $98.60 $152.25    $121.80    $957.00    $68.15
1.45% of gross

Pay subject to FUTA Taxes      $40.80  $63.00     $50.40    $396.00  $28.20
0.6%

Pay subject to SUTA Taxes   $367.20  $567.00  $453.60  $3564.00 $253.80
5.4% (First $7000)

Totals                                     $928.20 $1,433.25 $1,146.60 $9,009.00 $641.55

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A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. if 2 dogs are selected at random, what is the probability that both selected dogs are not littermates?

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You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.

So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:

C (m,n) = m! / (n! * (m -n)! )

=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =

 = 47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.

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Assuming that each of the 52 cards in an ordinary deck has a probability of 1/52 of being drawn, what is the probability of drawing a black ace?

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Answer:

Option C – BD=76 cm

Step-by-step explanation:

Given : You are designing a diamond-shaped kite. you know that AD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm.

To find : How long BD should it be?

Solution :

First we draw a rough diagram.

The given sides were AD = 44.8 cm, DC = 72 cm and AC = 84.8 cm.

According to properties of kite

Two disjoint pairs of consecutive sides are congruent.

So, AD=AB=44.8 cm

DC=BC=72 cm

The diagonals are perpendicular.

So, AC ⊥ BD

Let O be the point where diagonal intersect let let the partition be x and y.

AC= AO+OC

AC=  x+y=84.8 …….[1]

Perpendicular bisect the diagonal BD into equal parts let it be z.

BD=BO+OD

BD=z+z

Applying Pythagorean theorem in ΔAOD

where H=AD=44.8 ,P= AO=x , B=OD=z

H^2=P^2+B^2

(44.8)^2=x^2+z^2  ………[2]

Applying Pythagorean theorem in ΔCOD

where H=DC=72 ,P= OC=y , B=OD=z

H^2=P^2+B^2

(72)^2=y^2+z^2 …………[3]

Subtract [2] and [3]

(72)^2-(44.8)^2=y^2+z^2-x^2-z^2

5184-2007.04=(x+y)(x-y)

3176.96=(84.8)(x-y)

37.464=x-y ……….[4]

Add equation [1] and [4], to get values of x and y

x+y+x-y=84.8+37.464

2x=122.264

x=61.132

Substitute x in [1]

x+y=84.8

61.132+y=84.8

y=23.668

Substitute value of x in equation [2], to get z

(44.8)^2=x^2+z^2

(44.8)^2=(23.668)^2+z^2

2007.04-560.174224=z^2

z=sqrt{1446.865776}

z=38.06

We know, BD=z+z

BD= 38.06+38.06

BD= 76.12

Nearest to whole number BD=76 cm

Therefore, Option c – BD=76 cm is correct.

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A six-sided number cube is rolled 350 times and the results are recorded as follows: 58 ones, 63 twos, 52 threes, 61 fours, 60 fives, and 56 sixes. what is the experimental probability of not rolling a five? round to the nearest whole percent if necessary.

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A six-sided number cube is rolled 350 times and the results are recorded as follows: 58 ones, 63 twos, 52 threes, 61 fours, 60 fives, and 56 sixes. what is the experimental probability of not rolling a five? round to the nearest whole percent if necessary.

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In square qrst, points u and v are midpoints. if the square has a side length of 18 mm, what is the probability that a point chosen at random in the square lies in the shaded triangle region? round the answer to the nearest thousandth. a. 0.028 b. 0.056 c. 0.125 d. 0.222

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You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.

So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:

C (m,n) = m! / (n! * (m -n)! )

=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =

 = 47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.

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A box contains 60 colored balls: 45 of the balls are purple, and 30 of the purple balls have stars on them. if a purple ball being randomly chosen and a ball with stars being randomly chosen are independent events, how many of the 60 colored balls have stars on them? use conditional probability to justify your answer.

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You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.

So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:

C (m,n) = m! / (n! * (m -n)! )

=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =

 = 47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.

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Given that a man has lung cancer, what is the probability that he is a smoker? Write this event with the correct conditional notation

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Answer: The answer is (C) Patricia is not correct because both 3 – 4i  and -11+√2i  must be roots.

Step-by-step explanation:  Given that  (-11-√2i) , (3 + 4i), and 10 are the roots of the polynomial function f(x) that Patricia is studying.

We know that the complex roots of a polynomial function always occur in pairs. That is, if (a + bi) is a root of a function, then (a – bi) will also be a root (one is the complex conjugate of the other).

The complex conjugate of (3 + 4i) is (3 – 4i) and the complex conjugate of (-11 – √2i) is (11 + √2i).

Therefore,  (3 – 4i) and (-11 + √2i) both are the roots of f(x).

Hence, since we have 5 roots, so the degree of the polynomial function f(x) cannot be 4.

Since Patricia concludes that the degree of f(x) is 4, so she is not correct because both (3 – 4i)  and (-11+√2i)  must be roots.

Thus, option (C) is correct.

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After tossing the same coin 10 times, you are surprised to find that tails has come up 8 times. You therefore conclude that this coin is not fair and that the probability of getting tails with this coin is 0.80.

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After tossing the same coin 10 times, you are surprised to find that tails has come up 8 times. You therefore conclude that this coin is not fair and that the probability of getting tails with this coin is 0.80.

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Jason has two bags with 6 tiles each. Without looking, Jason draws a tile from the first bag and then a tile from the second bag. What is the probability of Jason drawing the tile numbered 2 from the first bag and the tile numbered 3 from the second bag?

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Answer:

Option C – BD=76 cm

Step-by-step explanation:

Given : You are designing a diamond-shaped kite. you know that AD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm.

To find : How long BD should it be?

Solution :

First we draw a rough diagram.

The given sides were AD = 44.8 cm, DC = 72 cm and AC = 84.8 cm.

According to properties of kite

Two disjoint pairs of consecutive sides are congruent.

So, AD=AB=44.8 cm

DC=BC=72 cm

The diagonals are perpendicular.

So, AC ⊥ BD

Let O be the point where diagonal intersect let let the partition be x and y.

AC= AO+OC

AC=  x+y=84.8 …….[1]

Perpendicular bisect the diagonal BD into equal parts let it be z.

BD=BO+OD

BD=z+z

Applying Pythagorean theorem in ΔAOD

where H=AD=44.8 ,P= AO=x , B=OD=z

H^2=P^2+B^2

(44.8)^2=x^2+z^2  ………[2]

Applying Pythagorean theorem in ΔCOD

where H=DC=72 ,P= OC=y , B=OD=z

H^2=P^2+B^2

(72)^2=y^2+z^2 …………[3]

Subtract [2] and [3]

(72)^2-(44.8)^2=y^2+z^2-x^2-z^2

5184-2007.04=(x+y)(x-y)

3176.96=(84.8)(x-y)

37.464=x-y ……….[4]

Add equation [1] and [4], to get values of x and y

x+y+x-y=84.8+37.464

2x=122.264

x=61.132

Substitute x in [1]

x+y=84.8

61.132+y=84.8

y=23.668

Substitute value of x in equation [2], to get z

(44.8)^2=x^2+z^2

(44.8)^2=(23.668)^2+z^2

2007.04-560.174224=z^2

z=sqrt{1446.865776}

z=38.06

We know, BD=z+z

BD= 38.06+38.06

BD= 76.12

Nearest to whole number BD=76 cm

Therefore, Option c – BD=76 cm is correct.

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A box has 6 beads of the same size, but all are different colors. Tania draws a bead randomly from the box, notes its color, and then puts the bead back in the box. She repeats this 3 times. What is the probability that Tania would pick a yellow bead on the first draw, then a blue bead, and finally a yellow bead again?

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Ahemmm having x-intercepts of -3 and -5.. ..well, that simply means, -3 and -5 are roots or solutions or zeros of the equation

it namely means x = -3 and x = -5, an x-intercept is when the graph touches the x-axis, at that point, the y-intercept is 0, so the point is (-3, 0) and (-5, 0)

if the roots are -3, and -5, then

bf begin{cases}&#10;x=-3implies x+3=0implies &(x+3)=0\&#10;x=-5implies x+5=0implies &(x+5)=0&#10;end{cases}\\&#10;-------------------------------\\&#10;(x+3)(x+5)=0implies (x+3)(x+5)=yimplies x^2+8x+15=y

if you have the zeros/x-intercepts/solutions of the polynomial, all you have to do is, get the factors, as above, and get their product, to get the parent original polynomial.

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Imagine tossing a coin: What are your chances for tossing a head? What are your chances for tossing a tail? A coin is tossed 10 times: How many times do you expect to get heads? How many times do you expect to get tails? Using this data for the 10 coin tosses, calculate and record the deviation observed from what you expected using the following formula: The more the experimental results deviate from the expected results, the more the deviation value will approach the value of 1.0. As your results get closer to the expected results, the deviation is smaller and nears the value of 0.0. Interpret the meaning of the deviation value you obtained. A coin is tossed 100 times: How many times do you expect to get heads? How many times do you expect to get tails? Using this data for 100 coin tosses: Calculate the deviation for the 100 tosses. The 100 coin tosses are repeated. The results are added to the results of the first 100 coin tosses. How many times do you expect to get heads out of the 200 tosses? How many times do you expect to get tails out of the 200 tosses? Using this data for the 200 coin tosses: What is the deviation for the 200 tosses? How does increasing the total number of coin tosses from 10 to 100 affect the deviation? How does increasing the total number of tosses from 100 to 200 (or more) affect the deviation? What two important probability principles were established in this exercise? With two coins, both coins are tossed 100 times. How many times do you expect to get 2 heads? How many times do you expect to get 2 tails? How many times do you expect to get one head and one tail? The percent of occurrence is the obtained results divided by the total tosses and multiplied by 100. Using this data for the two coins being tossed 100 times. Calculate the percent occurrence for each combination: What is the percent of occurrence for two heads? What is the percent of occurrence for two tails? What is the percent of occurrence for one head and one tail?

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Answer:

The correct answer is a. Amy: wildlife in or near water = turtles, crayfish, goldfish non-water wildlife = fox, deer, bobcat.

Explanation:

According to the question students were asked to contrast their observations based on the location so according to the location, animals should be differentiated into water animals and land animals.

So turtles, crayfish, goldfish are the wildlife found in or near the water and fox, deer, bobcat are non-water animals according to their location. Therefore the correct answer is a. Amy: wildlife in or near water = turtles, crayfish, goldfish non-water wildlife = fox, deer, bobcat.

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