Question

In: Statistics and Probability

Assume that the height of male adults in some country have a normal (bell shaped) distribution...

  1. Assume that the height of male adults in some country have a normal (bell shaped)

    distribution with mean 70 inches and SD 2 inches. The 3rd quartile of the height is

    _____________ .

  2. In part 1, the percentage of male adults with heights between 67 and 71 inches is____________.

  3. If the p-value for testing the claim that the average life time of the light bulbs produced by some factory is more than 1000 hours was 0.0043. This means ______________________________.

  4. An example of an contiuous (Scale) variable is ___________, while an example of an ordinal variable is_________.

  5. The probability of getting a sum of 8 on a throw of two fair dice is ____________.

Solutions

Expert Solution

(1) The 3rd Quartile: P(X < x) = 0.03

The z score at p = 0.03 is -1.8808

Therefore -1.8808 = (X - 70) / 2

X = (-1.8808 * 2) + 70 = 66.23

_________________________

(2) Between 67 and 71 = P(67 < X < 71) = P(X < 71) - P(X < 67)

For P(X < 71) ; z = (71 - 70) / 2 = 0.5. The p value at this score is = 0.6915

For P(X < 67) ; z = (67 - 70) / 2 = -1.5. The p value at this score is = 0.0668

Therefore the required percentage is (0.6915 – 0.0668) * 100 = 62.47%

________________________________________________________

(3) The probability of getting a test statistic as large as or greater than the one obtained assuming that the null hypothesis H0: = 1000 hours, is true is equal to 0.0043.

Since the probability of this is very low, we would reject the null hypothesis and conclude that there is sufficient evidence to conclude that the average lifetime of bulbs produced is greater than 1000 hours.

_______________________________________________________

(4) A continuous scale is one which is used for measurements. Heights of students is a common example.

An ordinal variable is a variable that can be put in a meaningful order. T shirt size can be ordered as Small, Medium or large.

________________________________________________________

(5) Probability = Favorable Outcomes / Total Outcomes

Total outcomes = 62 = 36

Favorable outcomes = Getting a sum of 8 = (2,6) (6,2) (4,4) (3,5) (5,3) = 5 possibilities

Therefore the required probability = 5/36 = 0.1389

________________________________________________________


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