January Regional Statistics Individual Solutions

  1. A – Skewed left distributions have a smaller mean than median.
  2. B – I is false, and SRS is always essential

II is true, it is one of the properties of the power of a test.

III is false, the probability of a Type I error is equal to the level, and is independent of the sample size.

  1. C – If A and B are independent, whether or not A occurs has no effect on the probability of B.
  2. C -

X / 1 / 2 / 3 / 4 / 5 / 6
P(X) / / / / / /

The triangular numbers in this distribution are 1, 3, and 6. The combined probability is .

  1. C – The area of a trapezoid is . The area of a density curve is always 1. Base 1 is 1, and Base 2 is 0.8-0.2=0.6. Solve for a, which is 1.25. The tenths digit of ln(a) is 2.
  2. D – The trapezoid must be divided into two triangles and a rectangle. The first triangle’s area is 0.125. Therefore, we must find the x value that would give the rectangle an area of (0.75-0.125) = 0.625.

0.625 = b*h, where the height is 1.25.

  1. C – There are 4 odd cards per suit: 3, 5, 7, and 9. Out of the red cards, there are 8 odd cards.
  2. A – This is a T test because we do not know the standard deviation of the population.

, , , and the P-value was 0.04095,

  1. A – There are 14 letters, and of those letters there are 4 letters that repeat: 3 O’s, 2 U’s, 2 L’s, and 2 D’s.

# of distinct permutations =

  1. D – P( Correctly calling Sophie her name | She is present)

To solve we must use the equation:

Sophie can either be alone, or with Joyce for her to be present. They are together 0.6 of the time, and Sophie is alone 0.25 of the time. Therefore, she is present 0.85 of the time.

To find P(correctly naming her present), you find 0.995*0.6 + 0.8*0.25, which is the total probability of correctly naming her, which comes out to 0.797. The final answer is 0.937647.

  1. C – We are finding: P(correctly naming Joyce | someone is called Joyce)

There are 4 ways someone can be called “Joyce”. Sophie can be incorrectly called “Joyce” when she is alone, Sophie and Joyce are together and they are correctly identified, Sophie and Joyce are together and they are incorrectly identified, and Joyce is alone and she is correctly called “Joyce”.

The sum of these probabilities is the probability that someone will be called “Joyce”.

This is: (0.25*0.2) + (0.6*0.995) + (0.6*0.005) + (0.15*0.8375) = 0.775625

The probability that Joyce is correctly named is:

(0.6*0.995) + (0.15*0.8375) = 0.722625

0.7226250.775625 = 0.931668

  1. C – The experimental units are the subjects in this experiment. In this case, the subjects suffering from a cold. The response variable is the variable being measured, which is the difference in temperatures.
  2. D – There are two factors: The solid medication and the liquid medication. There are 3 types of solid medication and 2 types of liquid medications. There are 6 possible treatments.
  3. B – The design is a simple randomized comparative design
  4. B – The line of random digits is divided into 2-digit numbers:

1387 38 15 98 95 05 29 05 08 73 59 2

13, 38, 15, 05, 29 are the first 5 numbers that are between 01 and 50 without repeats.

  1. C – Herschel misses the first free throw. There are two scenarios in which he makes the third free throw: Herschel misses the second free throw and makes the third, or he makes both the second and third free throws. Because he missed the first, his probability of success is halved to 0.3 for the second. If he misses the second, it is dropped to 0.15 for the third.

(1-0.3)*(0.15) + (0.3)*(0.3) = 0.195

  1. C – Brown Sugar misses the first free throw. There are two scenarios in which he makes the third free throw: He misses the second free throw and makes the third, or he makes both the second and third free throws. Because he misses the first, his probability of success does not change for the second. If he makes the second, his probability of success for the third increases to 0.65.

(1-0.6)*(0.6) + (0.6)*(0.65) = 0.63

  1. C – Venn diagram:

  1. B –
  2. A – Compare z-scores

Sebastian:

Jay:

Sebastian’s z-score is higher, so he did better than Jay.

  1. E – The LSRL is

However, the sum of the residuals is always 0.

  1. D – After plugging the points into a graphing calculator and obtaining the LSRL, the coefficient of determination is displayed as r2.
  2. B – Variance = = 4.2139

Standard deviation =

  1. B – The strongest correlation is the one where |r| is closest to 1.
  2. D – This is the only correct interpretation of the confidence interval.
  3. C – For the regression line , and
  4. A – The probability that Richard Pick wins is , the probability that Richard Morse Price II wins is , and the probability that Richard Farmer wins is .

P(Morse Price | PickC) = .

  1. A – Residual = . y = 14 at x=3. = -1 at x=3. (14 – (-1)) = 15.
  2. D –

, therefore

  1. C – Sum of (people who love it and break pencils) + (people who neither love nor hate it and break pencils) + (people who hate it and break pencils) = (0.25*0.001) + (0.4*0.05) + (0.35*0.2) = 0.09025