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GEN 305 Mathematical Concepts And Statistical Applications
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GEN 305 Mathematical Concepts And Statistical Applications
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Course Code: GEN 305
University: Purdue University Global
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Country: United States
Questions:
Gestation period The length of human pregnancies is approximately normally distributed with mean µ = 266 days and standard deviation σ = 16 days.
a.What is the probability a randomly selected pregnancy lasts less than 260 days?b.What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?c.What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?d.What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?
Answer:
Normal Distribution – Sampling Distributions – Central Limit Theorem
Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean inches and a standard deviation of inches. Compute the probability that a simple random sample of size results in a sample mean greater than 40 inches. That is, compute.
Solution
Let be the members of the sample with size, and
. If , follows a normal distribution with , so will also follows a normal distribution with mean 38.72 and standard deviation ( Hassett & Stewart, 2006)
. Here Z-score will be computed to facilitate the determination of probability from Z- tables (Hassett & Stewart, 2006, p.222). Z-score is computed as follows
Therefore, the probability will be computed as follows.
Therefore,
Gestation period: The length of human pregnancies is approximately normally distributed with mean µ = 266 days and a standard deviation of σ = 16 days.
Solution
The probability of a selected pregnancy lasts less than 260 days
From the question data
The probability of will be computed as follows
Francis (2004) stated that the probability of are obtained from the Standard Normal distribution table
The probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less
Solution
The probability to be computed is
Let be the members of the sample with size, and. According to Hassett & Stewart( 2006), If the length of pregnancy( ) is normally distributed with mean 266 and standard deviation 16, then will be normally distributed with mean 266 and standard deviation
Therefore,
The probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.
The probability to be computed is
Let be the members of the sample with size , and. Since the length of pregnancy is normally distributed with mean 266 and standard deviation 16, then will be normal with mean 266 and standard deviation ( Hassett & Stewart, 2006)
Therefore,
Probability a random sample of size 15 will have a mean gestation period within 10 days of the mean..
Solution
Let be the members of the sample with size , and. Since the length of pregnancy is normally distributed with mean 266 and standard deviation 16, then will be normal with mean 266 and standard deviation ( Hassett & Stewart, 2006).
First, computation of standard deviation in 10 days of the mean
Required to compute
Therefore, Probability a random sample of size 15 will have a mean gestation period within 10 days of the mean is 0.4448.
Problem: According to the U.S. Department of Agriculture, the mean calorie intake of males 20 to 39 years old, with standard deviation. Suppose a nutritionist analyzes a simple random sample of males between the ages of 20 and 39 years old and obtains a sample mean calorie intake of calories. Are the results of the survey unusual? Why?
The probability that a random sample of 35 males between the ages of 20 and 39 years old would result in a sample mean of 2750 calories or higher
The probability to be computed is
Let be the members of the sample with size, and. The calorie intake of male is assumed to be normally distributed with mean 2716 and standard deviation 72.8, then will be normally distributed with mean 2716 and standard deviation
The reason why results were unusual
The results of the sample are unusual due to fact that z-score (201.1467) obtained is beyond the limits whose probability can be read from the normal the Z-tables (Wackerly eta l., 2014)
Oil change: The shape of the distribution of time required to get an oil change at a 10-minute oil change facility is unknown. However, records indicate that the mean time for an oil change is µ = 11.4 minutes and standard deviation of σ = 3.2 minutes.
Solution
The required sample size for computing probabilities regarding the sample means using the normal model.
Francis (2004) stated that the sample size has to be large, greater than 30. “, is normally large”( Francis, 2004, p.480)
The probability that a random sample of oil changes results in a sample mean time less than 10 minutes.
The distribution of time for an oil change is normally distributed with mean 11.4 minutes and a standard deviation of 3.2 minutes since sample size ( ) is greater than 30. The probability to be computed, , since time is normally distributed, also the is normally distributed with mean 11.4 minutes and standard deviation (Hassett & Stewart, 2006)
Therefore,
Therefore,
Binomial Distributions – Normal Approximation to Binomial Distributions – Sampling Binomial Distributions
Fishing: Billfish Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In World Record Game Fishes (published by the International Game Fish Association), it was stated that in the Cozumel region, about 44% of strikes (while trolling) result in a catch. Suppose that on a given day a fleet of fishing boats got a total of 24 strikes. :
The probability that the number of fish caught was 12 or fewer
According to Ross (2009) the expectation of X in binomial distribution is given by
The variance of X is given by (Francis, 2004, p. 465)
Since the value of are greater than 5, the distribution will be approximated by normal distribution as follow (Francis, 2004).
Probability to be computed, =
Z-value is given
Therefore,
The probability that the number of fish caught was 5 or more
Considering the continuity correction factor, the distribution can be approximated by a normal distribution as follow.
Therefore,
The probability that the number of fish caught was between 5 and 12
From the ( b)and (c) above , and
Therefore,
Grocery Stores: New Products The Denver Post stated that 80% of all new products introduced in grocery stores fail (are taken off the market) within 2 years. If a grocery store chain introduces 66 new products.
Therefore, standard deviation
Since, the values of are greater than 5 then Continuity Correction factor will be utilized in approximating this distribution to normal one (Ross, 2006) .
The probability that within 2 years 47 or more fail
Considering the continuity correction factor
Therefore,
The probability that within 2 years 58 or fewer fail
Considering the continuity correction factor, the distribution can be approximated by a normal distribution ( Ross, 2009).
Therefore,
Thus
The probability that within 2 years 15 or more succeed
In this case, the mean be give
Considering the continuity correction factor, the distribution can be approximated by a normal distribution as follow.
Thus,
The probability that within 2 years fewer than 10 succeed
In this case, the mean is 13.2
The distribution can be approximated by normal distribution using continuity correction factor as follow.
Therefore,
Thus,
Crime: Murder: What are the chances that a person that is murdered actually knew the murderer? The answer to this question explains why a lot of police detective work begins with relatives and friends of the victim! About 64% of the people who are murdered actually knew the person who committed the murder. (Chances: Risk and Odds in Everyday Life by James Burke) Suppose that a detective file in New Orleans has 63 current unsolved murders. :
Since the values of are greater than 5, the distribution can be approximated by normal distribution using continuity correction factor as follow (Francis, 2004).
The probability that at least 35 of the victims knew their murderers
Considering the continuity correction factor, the distribution can be approximated by a normal distribution as follow
Therefore,
Thus,
The probability that at most 48 of the victims knew their murderers
Considering the continuity correction factor, the distribution can be approximated by a normal distribution as follow
Therefore,
Thus,
Basic Computation: distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
Suppose and. Can we approximate the distribution by a normal distribution? Why? What are the values of and
Solution
The distribution can be approximated by normal distribution since the values of and are greater than 5
The value of and value of
Suppose and . Can we safely approximate the distribution by a normal distribution? Why or why not?
Solution
The distribution cannot be safely approximated by normal distribution since the value of is not greater than 5 even though the value of is greater than 5. At the same time, sample size is less than 30(Francis 2004, p. 480).
Suppose and . Can we approximate the distribution by a normal distribution? Why? What are the values of and ?
Solution
Since the value of are greater than 5, distribution can be safely approximated by a normal distribution (Ross, 2009).
The values of and
Basic Computation: distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
Suppose and . Can we approximate the distribution by a normal distribution? Why? What are the values of and .
The distribution can be approximated by normal distribution since the values of and are greater than 5(Francis 2004).
The values of and
Suppose and . Can we safely approximate the distribution by a normal distribution? Why or why not?
Solution
The expectation of X in binomial distribution is given by (Ross, 2009, p.139)
The distribution cannot be safely approximated by normal distribution since the value of is not greater than 5. Even though the value of is greater than 5, the sample size is also less than 30.
Reference
Francis, A. (2004). Business mathematics and statistics. Cengage Learning EMEA.
Hassett, M. J., & Stewart, D. (2006). Probability for risk management. Actex Publications.
Ross, S. (2009). A First Course in Probability 8th Edition. Pearson.
Wackerly, D., Mendenhall, W., & Scheaffer, R. L. (2014). Mathematical statistics with applications. Cengage Learning.
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