If the midpoint of the line segment is AB has coordinates (3, -1) and A (-5,0) then: B=
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Michael Jordan's Freethrow, the population of the 6 countries and the distribution of the uninsured drivers are illustrations of probabilities
<h3>1. Michael Jordan Freethrow</h3>The given parameter is:
p = 83% -- the percentage of free throws he made
<u>Probability that he misses his 6th throw </u>
The probability that he misses his 6th throw is:
P = P(He made the first 5 throws) * P(He lost the 6th)
This gives
P = (83%)^5 * (1 - 83%)^1
P = 0.0670
<u>Expected number of miss</u>
The expected number till he misses the first free throw is:
E(x) = np
This gives
E(x) = 6 * 83%
E(x) = 4.98
Approximate
E(x) = 5
Hence, the probability that he misses his 6th throw is 0.0670 and the expected number till he misses the first free throw is 5
<h3>2. Country Populations</h3>The given parameters are:
Average population = 268.35 million
Standard deviation = 370.43 million
<u>The z-scores for the United States, Egypt, and Mexico</u>
This is calculated as:
z = (x - )/
So, we have:
US = (332 million - 268.35 million)/ 370.43 million
US = 0.17
Egypt = (104 million - 268.35 million)/ 370.43 million
Egypt = -0.44
Mexico = (128 million - 268.35 million)/ 370.43 million
Mexico = -0.38
<u>The countries from the smallest to the largest </u>
By comparing the z-score of the countries, the order of the countries are:
Egypt, Japan, Mexico, Brazil, United States and India
The countries are ranked that way because the higher the z-score, the higher the population
<h3>3. Uninsured drivers</h3>The given parameter is:
p = 12.6% -- the percentage of uninsured drivers
<u>Probability that exactly 5 drivers of 30 are uninsured</u>
This is calculated as:
P(x) = nCx * p^x * (1 - p)^(n-x)
So, we have:
P(5) = 30C5 * (12.6%)^5 * (1 - 12.6%)^(30-5)
Evaluate the product
P(5) = 0.1561
<u>Probability that the next uninsured driver is the 7th</u>
This is calculated as:
P = P(First 6 are insured) * P(7th is unisured)
This gives
P = (1 - 12.6%)^6 * (12.6%)^1
P = 0.0562
The probability that exactly 5 drivers of 30 are uninsured is 0.1561, and the probability that the next uninsured driver is the 7th is 0.0562
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