GEN 305 Mathematical Concepts And Statistical Applications

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

1. 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

1. 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,  

1. 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,  

1. 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?

1. 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  

1. 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

1. 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)

1. 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. :

1. 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, 

1. 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, 

1. 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) .

1. The probability that within 2 years 47 or more fail

Considering the continuity correction factor

Therefore,

1. 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

1. 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,          

1. 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).

1. 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,

1. 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.

1. 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  

1. 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). 

1. 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.

1. 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  

1. 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

1. Francis, A. (2004). Business mathematics and statistics. Cengage Learning EMEA.
2. Hassett, M. J., & Stewart, D. (2006). Probability for risk management. Actex Publications.
3. Ross, S. (2009). A First Course in Probability 8th Edition. Pearson.
4. Wackerly, D., Mendenhall, W., & Scheaffer, R. L. (2014). Mathematical statistics with applications. Cengage Learning.