∑f=80 Calculations The sample arithmetic mean x ̅=( ∑▒〖(MI)×f〗)/(∑▒f) =1535/80=19.19 The median x ̅ =〖 L〗_(1 )+( (N/2-∑▒〖fmed )×c〗)/fmed =15+((40-27))/((22))×5 = 15 + 2.95 = 17.95 〖 L〗_(1 )=Lower class limit of the median N=Total Frequency ∑▒█(fmed )=Sum of the frequencies of classes below the median class@) fmed =Frequency of the median class c=Width of the median class The mode (X ) ̂=〖 L〗_(1 )+ (f_(0-f_1 )/(2 f_0-f_(1 - ) f_2 ))×h =15+(22-19)/(2(22)-19-13)
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* These histogram are clearly asymmetrical and the superimposed normal curve is a little bit do not fit to them well‚ infact the histogram are skewed to the right so the sample is non-normal distribution * The sample mean is far above the median. Other important measures of the normality are the skewness and kurtosis. (In normal distribution the values of skew and kurtosis are close to 0). However in this sample‚ both of them are large positive. Moreover‚ we have z-score of skewness is 12
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specify case and variable? Q22: What are the necessary skills required to learn the SPSS? Q23: Give the name of any other data analysis software and explain their applications. Q24: What is a use of value option in variable view? Q25: Define Mean‚ Median and Mode Q26: What is correlation of coefficient? Q27: What is correlation of determination? Q28: What is test of significance? Q29: Explain different correlation value with scattered diagram. Q30: What is standard deviation? Q31: Define frequency
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Auditing: A Journal of Practice & Theory Vol. 29‚ No. 2 November 2010 pp. 115–140 American Accounting Association DOI: 10.2308/aud.2010.29.2.115 Do Abnormally High Audit Fees Impair Audit Quality? Jong-Hag Choi‚ Jeong-Bon Kim‚ and Yoonseok Zang SUMMARY: This study examines whether and how audit quality proxied by the magnitude of absolute discretionary accruals is associated with abnormal audit fees‚ that is‚ the difference between actual audit fee and the expected‚ normal level of audit
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variables. 5. The scores of the top ten finishers in a recent LPGA Valley of the Stars Tournament are listed below. (Source: Los Angeles Times) 73 67 72 67 76 72 73 68 71 73 Find the mode score. 73 6. Multiple choice: The Median is a. the most stable measurement of Central Tendency. b. the least stable measurement of Central Tendency. c. only
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length of its leaves. This data will then be used to analyse the tree’s ability to efficiently capture sunlight and move liquid around the tree. It is predicted that this report will prove that as a tree’s growing capacity increases the average‚ median and mean leaf size will decrease. This is likely due providing a more efficient means of transporting fluid around the tree. However to compensate for the shorter leaves’ having less ability to capture sunlight there will be an increased quantity
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Enclosure: 1. The report Acknowledgement In preparing report we thank the library manager of Dhaka Stock Exchange Mr. Attaur Rahman who supplied us with the “Monthly Review of DSE.”(A year end edition of 2003) and Library Manger of Security Exchange commission who assists us providing the information where we can collect required information. It was a great pleasure from our classmates from whom we received cordial guidance‚ suggestions for preparing this report specially our
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population. X = 567‚ X2 = 48‚165‚ N=7 Ordered array of data: 49‚ 63‚ 77‚ 85‚ 94‚ 98‚ 101 Median position = (n+1)/2 = 4‚ median is 85 Mean = X/N = 567/7 = 81kg 2 = (X– (X)/N) 2 /N = 319.7142857‚ = 17.88. Range = max-min = 101-49 = 52 4. Calories Fat 240 260 350 350 420 510 530 8 3.5 22 20 16 22 19 Descriptive stats for calories: Calories Mean 380 Standard Error 42.7618 Median 350 Mode 350 Standard Deviation 113.1371 Sample Variance 12800 Kurtosis -1.4478
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5.1 Z is the standard normal random variable it has a mean= 0 and sd = 1 (-10‚000 is lower bound‚ 10‚000 is upper bound if not given otherwise) Normal curve: The mean‚ median and mode are equal total areas=1 Symmetric and bell shaped never touches the x axis Z= Value-Mean/ standard deviation Properties of the standard normal distribution: cumulative area is close to 0 for z scores close to z= -3.49 cumulative area increases as the z scores increase cumulative area for z=0 is
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Discuss your 1st variable‚ using graphical‚ numerical summary and interpretation Numerical Summary of Credit Balance are as follows: Mean: 3970.5 Minimum: 1864 Standard Deviation: 931.9 Q1: 3109.3 Variance: 868429.8 Median: 4090 Skew: -0.15043 Q3: 4747.5 N: 50 Max: 5678 The histogram above shows the Credit Balance variable of the 50 customers surveyed. The histogram is almost symmetrical with one outlier which is the credit balance of $2‚000. While it
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