1. data:

1226651781762418211934

1082512642549317152517

A= mean = 474 / 26 = 237/13

B= median = 17

C= interquartile range =25-8 = 17

D= the number of outliers in the data set 1.5(17)=25.5; 25+25.5=50.5 , so 54 is an outlier. = 1

2. A density curve consists of a straight line segment that begins at the origin and has a slope of 1.

A. The x-coordinate of the right endpoint of the segment= so that the area underneath the curve will be 1. =

B. Median of the density curve =

C. IQR of the density curve =

D. So, since x and y are both 0.5, part D is

3. The following are the percent of each grade on last year’s AP Statistics class:

Grade 1 2 3 4 5

% .21.14.30.17.18

A: the mean of the distribution=(1)(.21)+(2)(.14)+3(.30)+4(.17)+5(.18)=2.97

B: the variance of the distribution =

4.

A: The probability of drawing a red card or an ace= 28/52

B: The probability of drawing a red ace = 2/52

C: Draw three cards, one at a time and without replacement. Probability that all three cards are the same suit=

5. The height of female students at Vero Beach High School is measured in inches, andthe data forms a normal distribution. The mean of the data is 62 inches with a standarddeviation of 4 inches. Use the information to answer the following parts. Each part isindependent of every other part. Round each probability to the nearest hundredth.

A) Find the probability that a female student is taller than 70 inches = normalcdf(70,999,62,4)=0.02

B) Find the probability that a female student is shorter than 59 inches=normalcdf(-999,59,62,4)=0.23

C) Find the probability that a female student has a height between 59 and 70 inches=normalcdf(59,70,62,4)=0.75

D) Find the probability that a female student is taller than 72 inches or shorter than 54inches=

0.00621+0.02275=0.03

A – B + C – D=0.02-0.23+0.75-0.03=0.51

6. A. p=140/360=7/18; geompdf(7/18,11)=0.0028

B. binompdf(5,220/360,4)=0.2712

A+B=0.274

7.

X) P(C | M) =24/88 = 3/11

Y) P(BandF) =8/127

Z) P(M | P') = 53/76

X Y Z=

8. ={All Possible Values of Probability}-{All Possible Values of Correlation} =

A = {All Possible Values for Probability}

B = {All Possible Values for Correlation}

C = {All Real Numbers}

9.

A = Median salary = 21,000

B = Range of Salaries =61,000

C = Sum of the Values of Outliers

IQR=18,000

1.5IQR=27,000

33,000+27,000=60,000

Outliers are 71,000 and 72,000 so the sum is 143,000

C-A+B =143,000-21,000+61,000=183,000

10. Linreg(cost,calories):

Let A = Slope of the Regression Line to the nearest tenth =3.9

Let B = y coordinate of the Y-Intercept of the Regression Line to the nearest whole number =329

Let C = Correlation between Calories vs. Cost to the nearest hundredths place=0.05

Let D% = Coefficient of Determination between Calories vs. Cost to the nearest integer=0

A+B+C-D=3.9+329+0.05-0=332.95

11.{the positive integral factors of 120}={1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120}

A) the mean of set X=22.5

B) the median of set X=11

C) the interquartile range of set X=22.5

D) the variance of set X=933

ABCD=5195643.75

12. x=72; z-score is 1.4669 (found by using invnorm(1-.0712)); so A=72-1.46695B

X=62; z-score is -2.0355 (found by using invnorm(.0209)); so A=62+2.0355B

A = the mean of the distribution =67.81135

B = the standard deviation of the distribution=2.855

A – B=64.96

13.

14. Q3 =25

1226651781762418211934

1082512642549317152517

Answers

  1. 0.51
  2. 0.274
  3. Φ (null set)
  4. 183,000
  5. 332.95
  6. 5195643.75
  7. 64.96
  8. 10
  9. 25