## Mr. Felix K. Colon

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QUIZ 1:  VARIATION AND CORRELATION

1.     (35 POINTS) The number of hours, h, it takes for a block of ice to melt varies inversely as the temperature, t.  If it takes 2 hours for a square inch of ice to melt at 65º, how many hours (to the nearest tenth) does it take to melt the same block of ice when the temperature is 80?

h1  t1 = h2 t2

2 hrs x 65 degrees = h2 x 80 degrees

130 = h x 80

130/80 = h

1.625   hrs. = h

It takes approximately 1.6 hrs. to melt the same block of ice when the temperature is 80 degrees.

2.     (15 points) The weight of an object varies inversely with the square of its distance from the center of the earth.  Assume that the radius of the earth is 3960 miles.  If an astronaut weights 180 pounds on the surface of the earth, what does he weigh when traveling 300 miles above the surface of the earth?

w1 x (d2)1 =  w2 x (d2)2 ; 3960 + 300 = 4260 miles from the center of the earth.

180 x 3960^2 = w2 x 4260^2

180(3960^2)/4260^2 = (w x 4260^2)/4260^2

154.54 lbs. = w

He weighs approximately 156 lbs. when traveling 300 miles above the surface of the earth.

(5 points each) Mark T (true) or F (false):

3.     Correlation coefficient does not have a range from -1 to 1. __F___

4.     A “r” value of less than -0.8  indicates a weak correlation.  ___F____

5.     A strong correlation shows cause and effect. ___F____

6.     A “r” value of 0.3 has a tight cluster of points around the line of best fit. __F___

7.     The points on the line of best fit represent the predicted outcomes. ___T____

(25 points) Sketch a scatterplot whose “r” value is -0.99

QUIZ 2:  QUARTILES, MODE, & MEDIAN

1. Which can be considered examples of a data set that is bimodal (circle):

a.  1,2,3,3,4,5, 6,6       b.  1,2,3,3,3,4,5,6,6,6              c. 1,1,2,3,4,4,5,5,6,7,8,9,9

A and B are bimodal.

2. What is another name for Q2?

Another name for Q2 is median.

3. What does it mean that Q1 = 72?

25% of the values are less than 72.

4. What does it mean that Q2 = 84?

50% of the values are less than 84.

5. What does it mean that Q3 = 89?

75% of the values are less than 89.

QUIZ 3: NORMAL DISTRIBUTION & STATISTICAL MEASUREMENTS

PART 1 (20 POINTS EACH):

1.     On a standardized national exam, the mean score was 90.5 and the standard deviation was 16. What is the percentile, to the nearest percent, of a student who scored 66.5?

90.5 = , 16 =

90.5 – 66.5 = 24

24/16 = 1.5

66.5 is located – 1.5 from the mean.  Adding percentages to the left of -1.5 we get 0.1 + 0.5 + 1.7 + 4.4 = 6.7%.  Therefore, 66.5 is at the 6.7 percentile.

2.     On a standardized national exam, the mean score was 128.9 and the standard deviation was 15.1. What percentage of the students scored between 98.7 and 159.1?

128.9 = , 15.1 =

128.9 – 98.7 = 30.2  à        30.2 / 15.1 =  2

159.1 – 128.9 = 30.2   à     30.2 / 15.1 =  2

The scores between 98.7 and 128.9 is between -2 and 2.

Adding the percentages between -2 and 2 we get 95.4% of the students scoring between 98.7 and 159.1.

3.     On a standardized national exam, the mean score was 82.4 and the standard deviation was 8.6. If 6840 people took the exam, approximately how many people scored higher than 91?

82.4 = , 8.6 =

91 – 82.4 = 8.6  à  8.6 / 8.6   =

If we add the percentages on the normal curve higher than 1  , we get 9.2 + 4.4 + 1.7 + 0.5 + 0.1 = 15.9%.  Therefore, .159(6840) = 1087.56  1088 people scored higher than 91.

4.     On a standardized national exam, the mean score was 168.8 and the standard deviation was 14.2. If 6950 people took the exam, approximately how many people scored less than 140.4?

168.8 = , 14.2 =

168.8 – 140.4 = 28.4  à 28.4 / 14.2 = 2

If we add the percentages on the normal curve lower than 2 , we get 0.1 + 0.5  + 1.7 = 2.3%.  Therefore, .023(6950) = 159.85  160 people scored less than 140.4.

QUIZ 3:  PART 2 (5 POINTS EACH):

5.     An example of measurement of position:

a.  range                     b. Q3                           c. mean

Q3 is a measurement of position

6.   An example of a measurement of central tendency:

a. Q1                           b. variance                 c.. median

The median is a measurement of central tendency.

7.    An example of a measurement of dispersion:

a.  mode                      b. maximum               c. standard deviation

A standard deviation is a measurement of dispersion.

8.    Interquartile range (IQR) is equal to:

a.  50% of the data               b. the middle 50% of the data

c.  the max – the minimum

The middle 50% of the data is called the interquartile range  (IQR).

QUIZ 4: PERMUTATION, COMBINATION, COUNTING PRINCIPLE

PART I (20 POINTS EACH)

1. What are the possible arrangements for …?

a.     2 ties,  5 pants, 3 shirts

2 x 5 x 3 = 30

2. What are the possible arrangements for …..?

a.     1st, 2nd, and 3rd for a race with 5 runners

nPr = 5P3 = 5!/(5-3)! = 5x4x3x2x1/2x1 = 60

3.What are the possible arrangements for a committee of 3 employees out of 7 to discuss issues at a job?

nCr = 7C3 = 7!/3!(7-3)! = 7x6x5x4x3x2x1/3x2x1(4x3x2x1) = 7x6x5/3x2x1 = 35

4. A committee must be formed with 3 teachers and 6 students. If there are 6 teachers to choose from, and 10 students, how many different ways could the committee be made?

6C3 x 10C6

= 6x5x4x3x2x1/3x2x1(3x2x1) x 10x9x8x7x6x5x4x3x2x1/ 6x5x4x3x2x1(4x3x2x1)

= 6x5x4/3x2x1            x          10x9x8x7/4x3x2x1 = 20         x          10x3x7  =  4200

PART II (5 POINTS EACH)

5.     For each example, indicate what concept applies, multiplication counting principle (M), permutation (P), or combination (C).

a.     ___c_____ a committee of 6 students out of 20

b.     ____p____ a race of 9 runners who are to be given awards for 1st, 2nd, 3rd

c.     ____m____ the different number of (one cheese and one meat) sandwiches one can make with 3 different cheeses and 2 different meats.

d.     ____c____ a team of 3 students out of 12 to play another team.

QUIZ 5:  BINOMIAL PROBABILITY: EXACT, AT LEAST, AT MOST

PROBLEMS 1-3 WORTH 28 POINTS EACH:

1.     COLON wants to go camping in a Florida state park where the probability of rain on any given day is 68%. What is the probability that it will rain on at least five of the six days I am there? Round your answer to 4 decimal places.

6C5  .68^5  .32^1     =          .2791552451

6C6  .68^6  .32^0     =          .0988674826

total    =          .3780227277

The probability that it will rain on at least five of the six days to 4 decimal places is .3780 or 37.80%.

2.     The probability that the Lights Out basketball team wins a game is 80%.What is the probability that they win exactly 3 of the next 5 games?  Round your answer to 4 decimal places.

5C3  .80^3  .20^2       =          .2048

The probability that they win exactly 3 of the next 5 games is .2048 or 20.48%.

3.     A coin is tossed 5 times.  What is the probability that it lands tails at most 2 out of 5 times?

5C0  .5^0  .5^5           =          .03125

5C1  .5^1  .5^4           =          .15625

5C2  .5^2  .5^3           =          .31250

total     =          .5000

The probability that it lands tails at most 2 out of 5 times is .5000 or 50.00%.

4.     True or False (4 points each)   For a binomial probability experiment:

a.     A binomial experiment has only 2 possible outcomes.      ____T____

b.     q = 1 + p                                                                           ____F____

c.     In a binomial probability experiment, the outcome of one event does not affect the probability of the outcome of another trial or attempt. ___T____

d.     In a binomial experiment, nCr is used because the order of successes is important.                                                                                   ____F____

TEST:  Exponents, Depreciation, Linear Regression, Exponential Growth,  Compound Interest, and Half-Life

1.  (10 POINTS)   Simplify:

= 5(x^3)y/(3^-2)(x^-8)(y^-2)

= (3^2)(5)(x^3)(x^8)(y)(y^2)

= 45(x^11)(y^3)

PROBLEMS 2-6 (18 POINTS EACH)

Depreciation: y = a(1-r)t

Compound Interest A=P(1+r/n)nt

Half-life: y = a(.5)t/h

2.     A Toyota Avalon car depreciates at 9.2% per year.

a.     What is the value of the car after 5 years if the car was purchased for \$44,000?

Y = a(1-r)^t

Y = 44000(1-0.092)^5  \$27157.01

The value of the car after 5 years, if the car was purchased for \$44000, is approximately \$27157.01

b.     In the future, the same car is valued at \$33,000.  How old do you expect the car to be? (Based on your calculations.)

33000 = 44000(1-.092)^t

33000/44000 = 44000(1-.092)^t / 44000

.75 = .908^t

log .75 = t log .908

log.75/log .908 = (t log .908)/ log.908

2.980824667 = t

A car valued \$33000 that was purchased originally for \$44000 is approximately 3 years old.

3.     A linear regression was calculated to be y = ..567x + 12.389 (use 3 places)

a.     What is predicted when x = 3?

Y = .567x + 12.389

Y = .567(3) + 12.389 = 14.090

TEST:  Exponents, Depreciation, Regression,  Compound Interest, and Half-Life

3.     b. What is x when y = 15.224?

15.224 = .567x + 12.389

15.224 – 12.389 = .567x + 12.389 – 12.389

2.835 = .567x

2.835/.567 = .567x / .567

5 = x

4.     An exponential regression was calculated to be y = 8.541(2.314)x

a.     What is predicted for year 1994 if x = 0 corresponds to 1990? (3 places)

If x = 0 for 1990, then x = 4 for 1994.  Therefore,

Y = 8.541(2.314)^4

Y = 244.8849754  to 3 places approximately equals 244.885

b.     What is x when y = 7021.268? (nearest integer)

7021.268 = 8.541(2.314)^x

7021.268/8.541 = 8.541(2.314)^x / 8.541

log (7021.268/8.541) = x log 2.314

log (7021.268/8.541) / log (2.314) = x log 2.314 / log 2.314

7.999999977 = x  and to the nearest integer 8 = x

5.     If a bank offers 1.4% interest and compounds daily and you deposit \$8000 and leave it untouched, how much money would you have after 10 years?

A=P(1+r/n)^(nt)

A = 8000(1 + 0.14/365)^(365x10)

A = 9202.165685

After 10 years, one would have approximately \$9202.17

6.     Element X is an isotope.  Solve for the missing variable:

Future Amt.    Initial Amount           time                half-life

35                                1120                                              5 yrs

y = a(.5)^(t/h)

35 = 1120(.5)^(t/5)

35/1120 = 1120(.5)^(t/5) / 1120

Log (35/1120) = t/5 log .5

Log (35/1120) / log (.5) = (t/5) log .5 / log .5

5 = t/5

5 (5) = (t/5) 5

25 = t

It takes 25 years for element X to decay to 35 gms.