Directions: Use your knowledge of the Poisson Distribution to answer Questions 1

Question

Directions: Use your knowledge of the Poisson Distribution to answer Questions 1 through 3.

1. Based on data from the U.S. National Center for Health Statistics, 11 babies are born in the village of Westport each year.

a. Find the mean number of births per day.

b. Find the probability that on a given day, there are no births.

c. Find the probability that on a given day, there is at least one birth.

2. Dandelions are studied for their effects on crop production and lawn growth. In one region, the average number of dandelions per square meter was found to be 7 (based on data from Manitoba Agriculture and Food).

a. Find the probability of at most 2 dandelions in an area of 1 square meter.

3. According to the World Almanac, for a recent period of 100 years, there were 93 major earthquakes (at least 6.0 on the Richter scale).

a. Find the probability that the number of earthquakes in a randomly selected year is 4.

Directions: Use the Standard Normal Distribution to answer Questions 4 through 6.

4. Using the standard normal tables, determine the following probabilities. Also, for each item, provide a sketch of the associated areas.

a. P(0< z <1.24) b. P(z>3.22) c. P(z < -3.39) d. P(-2.64 < z < 3.32)

5. If we assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15:

a. Find the probability that a randomly selected adult has an IQ greater than 131.5 (the requirement for membership in the MENSA organization).

b. Find the probability that a randomly selected adult has an IQ between 110 and 120.

6. The length of pregnancies is normally distributed with a mean of 268 days and a standard deviation of 15 days.

a. One classic use of the normal distribution is inspired by a letter to "Dear Abby" in which a wife claimed to have given birth 308 days after a brief visit from her husband, who was serving in the Navy. Given this information, find the probability of a pregnancy lasting longer than 308 days. What does the result suggest?

Explanation / Answer

2)

f(0;7) = 1e^-7 / 1 = 0.000911882

5)Well there are ways to calculate these. Or you can use some common sense.
15. With a mean of 100 for anormal distribution, there is 50% chance for <100. So for <131.5, there'd be an even larger chance. Only one answer has a % larger than 50%. So B.

6)

308 - 265 = 43 days. With astandard deviationof 15
She is 43/15 standarddeviationsover the mean.
2.87 standard deviations puts her in the 99.8 percentile. Which means there are .2% of the population have agestation periodas long or longer. Rare yes, but not impossible.
To put it into perspective there are 200,000 babies born every day, multiplying there are 400 of these children with the same or longer gestation.

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