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PART A:

Consumers spend an average of \$21 per week in cash without being aware of where it goes (data extracted from ‘Snapshots: A Hole in Our Pockets,’ USA Today, January 18, 2010, p. 1A). Assume that the amount of cash spent without being aware of where it goes is normally distributed and that the standard deviation is \$5.

(a) What is the probability that a randomly selected person will spend more than \$25?

(b)  What is the probability that a randomly selected person will spend between \$10 and \$20?

(c)  Between what two values will the middle 95% of the amounts of cash spent fall?

PART B:

The file Property Taxes contains the property taxes per capita for the 50 states and the District of Columbia. Decide whether the data appear to be approximately normally distributed.

Decide whether the data appear to be approximately normally distributed by:

(a) Constructing a box plot.

(b) Constructing a histogram.

(c) Comparing data characteristics to theoretical properties.

(d) Constructing a Quantile-Quantile Normal Probability Plot.

You must include all Excel results for this data as part of the justification for your conclusions. (Include them within the appropriate section – not at the end of your document.)

Your assignment must be submitted as a single word document via LMS.  You will find the data in Property Taxes in LMS under Topic 5, Data Files.1 Based on Question 6.18 in Levine et al (2013)

PART A: Completed questions.

• An introduction, where the key aspects of this topic are fully discussed. (This discussion should be referenced using the Chicago referencing method)
• The body of your assignment, where all of your findings are presented. (This should include relevant Excel results, graphs, tables) Each of the pieces of information that are included need to be discussed in relation to the given question.
• A Bibliography which includes all resources referenced within your assignment.

Marks will be allocated according to the quality of each of the components above. It is important that you present your assignment with sufficient detail to fully answer the given question, while at the same time present your arguments in a concise way. (As a guide: Word Limit, at least 1500 words).

## Probability of Cash Spending

Probability that a randomly selected person would spend more than \$25

The requisite probability would be the area under the blue section of the curve.

Therefore, the probability that a randomly selected person would spend more than \$25 is 0.2119.

• Probability that a selected person would spend between \$10 and \$20

The requisite probability would be the area under the blue section of the curve.

Therefore, probability that a selected person would spend between \$10 and \$20 is 0.4068.

• The middle 95% is indicated by the 0.95 area about the mean value. This represents that 0.95/2 = 0.4750 area would fall on the left side of the mean and also the 0.4750 area would fall on the right side of the mean.

Therefore, 95% of the total amount of cash spent lie between the range \$11.8 and \$30.8.

The data file regarding prevalent property taxes per capita in the various states has been presented. The objective of this report is to analyse the given data and determine if it is appropriate to conclude that the given data is normally distributed or not. This may be served in a plethora of ways such as through boxplot, histogram or normal probability plot construction (Bryc 2012, 96). Additionally, it may also be possible to compare whether the given data tends to comply with the various characteristics that are usually associated with any data that tends to follow a normal distribution (Elzey 2001, 79).  One of the properties in this regard is that the various measures of central tendency must coincide. This is related to the symmetric nature of the normal curve which is also indicative of the skew being zero (Bearver 2012, 95). Presence of skew leads to tail either on the right or left which in turn would highlight the non-normality of the concerned data (Bulmer 2012, 89). Additionally, there are certain theoretical properties particularly in relation to the particular distribution of data which must be exhibited which is expected from a normal distribution. Hence, comparison of the actual properties of the given data with the expected theoretical properties may be carried out which would enable in understanding whether normality persists or not for the given data (Weiers 2010, 115). The findings of the various techniques applied would be concluded at the end along with the underlying distribution of the variable of interest i.e. property tax per capita.

Analysis

For analysing the given data, various techniques have been implemented and presented below along with the potential implications of the same for normality.

• Construction of box plot

On the basis of the given boxplot along with the five number summary, it is apparent that the above boxplot has a tail on the right side or the higher side. This is representative of the presence of positive skew. However, for a normal distribution the skew should be zero. The non-zero value of skew is representative of the given distribution not being normal. Thus, based on the boxplot, it is apparent that the given data is nor normally distributed.

• Histogram

The histogram indicated above is not symmetric as it is skewed towards the right which is apparent from the presence of a long tail towards the right. Thus, it seems that there are certain extremely high values which are part of the given data which are contributing to the skew. The presence of skew indicates the non-normality in the data as for a normal distribution, the skew is expected to be zero. Thus, based on the above histogram, it is apparent that the given data is nor normally distributed.

• Summary statistics for the variable property taxes per capital (\$)

## Box Plot Analysis

From the above summary statistics, it is apparent that the mean value is higher than the median. This is on account of positive skew present in the data. The given data contains some exceptionally high values which are having a distorting effect on the mean which is pushing the value higher. However, the median is insulated from such extreme values and represented an undistorted view. The difference in mean and median values is also indicative of the non-normality of the given data.

From the above, it is apparent that there does not seem to be a significant deviation between the IQR (Inter Quartile Range) and 1.33 times standard deviation based on which, it would be correct that the given data is approximately normal.

Significant deviation is observed between the range and six times the standard deviation. This violates the expected theoretical property of a normal distribution.

Mean = \$ 1,040.86

Standard Deviation = \$ 428.54

Mean + 1*Standard Deviation = 1040.86 + 428.54 = \$ 1,469.40

Mean - 1*Standard Deviation = 1040.86 - 428.54 = \$ 612.32

Total number of values in the given data = 51

Number of values lying between Mean +/- one standard deviation = 32

Percentage of values lying between Mean +/- one standard deviation = (32/51)*100 = 62.75%

Mean + 1.28*Standard Deviation = 1040.86 + 1.28*428.54 = \$ 1,589.39

Mean – 1.28*Standard Deviation = 1040.86 – 1.28*428.54 = \$ 492.33

Total number of values in the given data = 51

Number of values lying between Mean +/- 1.28 standard deviation = 40

Percentage of values lying between Mean +/- 1.28 standard deviation = (40/51)*100 = 78.43%

Mean + 2*Standard Deviation = 1040.86 + 2*428.54 = \$ 1,897.94

Mean - 2*Standard Deviation = 1040.86 – 2*428.54 = \$ 183.79

Total number of values in the given data = 51

Number of values lying between Mean +/- two standard deviation = 48

Percentage of values lying between Mean +/- two standard deviation = (48/51)*100 = 94.12%

It is apparent from the above distributions that the theoretical expectations are not fulfilled as the values that are lying within the defined intervals tend to be lower than the expected percentage. This may be on the account of certain values which are extremely high and thus lying at the right end and hence not being included in the above computation which is responsible for the shortfall observed.

Also, the skewness of the given data is 0.60. The positive value of the skew confirms the presence of the tail on the right side that was inferred from the boxplot and also the histogram. Further, the kurtosis of the given data is also -0.11 which is different from the value expected for a data distributed normally.

It is apparent from the above observations that the given data is not normal as neither of the theoretical properties seem to be fulfilled except barring the first one as per which IQR is 1.33 times the standard deviation. Hence, it would be appropriate to conclude that the given data is not normally distributed.

• Normal probability plot

The relevant normal probability plot for the given data is indicated below.

The respective table used to derive the normal probability plot highlighted above is indicated below.

The normality probability plot for a variable that is normally distributed is expected to be linear in nature. However, in the given case, it is apparent that the normal distribution plot has a lot of breaks and is not linear but it having a mild curve due to which there is a S shape distribution. Further, there is presence of outliers especially on the higher side which have been confirmed through other means as well. Based on the available evidence, the non-normality of the variable at hand is reflected even from the given measure also.

Conclusion

The above analysis deployed various techniques in order to ascertain the underlying distribution of the given data so as to opine whether it is normal or not. Based on the boxplot and the histogram, it is apparent that the data is skewed towards the right which implies that the data would have a non-normal distribution. Further, based on the theoretical properties which are associated with a normal distribution, it is also apparent that the given data does not represent a normal distribution hence majority of the properties that a normal distribution would comply with are not being complied with the current data. Additionally, the summary statistics also hinted to the non-coincidence of central tendency measures. Also, the probability distribution plot is representative of the non-normality of the given data. Thus, based on the available evidence, it would be fair to conclude that the given data is not normally distributed.

References

Bryc, Wlodzimierz. The Normal Distribution: Characterizations with Applications. London: Springer Science & Business Media, 2012.

Bearver, Barbara. Introduction to Probability and Statistics. Sydney: Cengage Learning, 2012.

Elzey, Freeman. Elementary Statistical Techniques. Melbourne: Brooks, 2001.

Bulmer. Principles of Statistics. New York: Courier Corporation, 2012.

Weiers, Ronald. Introduction to Business Statistics. Sydney: Cengage Learning, 2010.

Cite This Work

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