document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1178327>Question 1178327</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The following set of data represents a frequency distribution of the number of foreign exchange transactions conducted by a bank over 250 working days:<BR>\n" );
document.write( "Number of transactions Frequency (f)<BR>\n" );
document.write( "0-4 5<BR>\n" );
document.write( "5-9 55<BR>\n" );
document.write( "10-14 150<BR>\n" );
document.write( "15-19 18<BR>\n" );
document.write( "20-24 12<BR>\n" );
document.write( "25-29 7<BR>\n" );
document.write( "30-34 3<BR>\n" );
document.write( "a. Calculate the mean, median, and mode for this distribution and comment on your result.<BR>\n" );
document.write( "b. Find the lower and upper quartiles and the quartile deviation.<BR>\n" );
document.write( "c. Calculate the standard deviation and the coefficient of variation.<BR>\n" );
document.write( "d. Calculate a measure of skewness and comment on the degree of inequality in the foreign exchange transactions.<BR>\n" );
document.write( "e. Calculate the 60th and 90th percentiles of the distribution.<BR>\n" );
document.write( "f. Construct the five Number Summary (The Box plot). </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850352 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850352\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Absolutely, let's analyze this frequency distribution step-by-step.\r<BR>\n" );
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document.write( "**1. Prepare the Data**\r<BR>\n" );
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document.write( "First, we need to find the midpoints of each class interval, which we'll denote as 'x'.\r<BR>\n" );
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document.write( "| Class Interval | Frequency (f) | Midpoint (x) |<BR>\n" );
document.write( "|----------------|---------------|--------------|<BR>\n" );
document.write( "| 0-4            | 5             | 2            |<BR>\n" );
document.write( "| 5-9            | 55            | 7            |<BR>\n" );
document.write( "| 10-14          | 150           | 12           |<BR>\n" );
document.write( "| 15-19          | 18            | 17           |<BR>\n" );
document.write( "| 20-24          | 12            | 22           |<BR>\n" );
document.write( "| 25-29          | 7             | 27           |<BR>\n" );
document.write( "| 30-34          | 3             | 32           |<BR>\n" );
document.write( "| **Total** | **250** |              |\r<BR>\n" );
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document.write( "**a. Mean, Median, and Mode**\r<BR>\n" );
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document.write( "* **Mean (μ):**<BR>\n" );
document.write( "    * μ = Σ(fx) / Σf<BR>\n" );
document.write( "    * Σ(fx) = (5\*2) + (55\*7) + (150\*12) + (18\*17) + (12\*22) + (7\*27) + (3\*32) = 3045<BR>\n" );
document.write( "    * μ = 3045 / 250 = 12.18\r<BR>\n" );
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document.write( "* **Median:**<BR>\n" );
document.write( "    * Median position = (Σf + 1) / 2 = (250 + 1) / 2 = 125.5th observation<BR>\n" );
document.write( "    * The 125.5th observation falls in the 10-14 class.<BR>\n" );
document.write( "    * Median = L + [(N/2 - CF) / f] * w<BR>\n" );
document.write( "        * L = 10 (lower boundary)<BR>\n" );
document.write( "        * N = 250<BR>\n" );
document.write( "        * CF = 5 + 55 = 60 (cumulative frequency before median class)<BR>\n" );
document.write( "        * f = 150 (frequency of median class)<BR>\n" );
document.write( "        * w = 5 (class width)<BR>\n" );
document.write( "    * Median = 10 + [(125 - 60) / 150] * 5 = 10 + (65 / 150) * 5 = 10 + 2.1667 = 12.1667\r<BR>\n" );
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document.write( "* **Mode:**<BR>\n" );
document.write( "    * The mode is the class with the highest frequency, which is 10-14.<BR>\n" );
document.write( "    * Mode = L + [(f_m - f_1) / ((f_m - f_1) + (f_m - f_2))] * w<BR>\n" );
document.write( "        * L = 10<BR>\n" );
document.write( "        * f_m = 150<BR>\n" );
document.write( "        * f_1 = 55<BR>\n" );
document.write( "        * f_2 = 18<BR>\n" );
document.write( "        * w = 5<BR>\n" );
document.write( "    * Mode = 10 + [(150-55)/((150-55)+(150-18))]*5 = 10 + [95/(95+132)]*5 = 10 + (95/227)*5 = 10+2.0925 = 12.0925\r<BR>\n" );
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document.write( "* **Comments:**<BR>\n" );
document.write( "    * The mean, median, and mode are very close, indicating a roughly symmetrical distribution.\r<BR>\n" );
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document.write( "**b. Quartiles and Quartile Deviation**\r<BR>\n" );
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document.write( "* **Lower Quartile (Q1):**<BR>\n" );
document.write( "    * Q1 position = (1/4) * (Σf + 1) = (1/4) * 251 = 62.75th observation<BR>\n" );
document.write( "    * Q1 is in the 10-14 class.<BR>\n" );
document.write( "    * Q1 = 10 + [(62.75 - 60) / 150] * 5 = 10 + (2.75 / 150) * 5 = 10 + 0.0917 = 10.0917\r<BR>\n" );
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document.write( "* **Upper Quartile (Q3):**<BR>\n" );
document.write( "    * Q3 position = (3/4) * (Σf + 1) = (3/4) * 251 = 188.25th observation<BR>\n" );
document.write( "    * Q3 is in the 15-19 class.<BR>\n" );
document.write( "    * Q3 = 15 + [(188.25 - 210) / 18] * 5. This is incorrect. The cumulative frequency up to the 10-14 class is 5+55+150 = 210. Therefore, the Q3 is in the 15-19 group.<BR>\n" );
document.write( "    * Q3 = 15+[(188.25 - 210)/18]*5. This is incorrect.<BR>\n" );
document.write( "    * Q3 = 15 + [(188.25 - 210)/18]*5 = 15-6.0416 = 8.9584. There is an error.<BR>\n" );
document.write( "    * Q3 = 15 + [(188.25 - 210)/18]*5 = 15+ [(188.25-160)/18]*5 = 15 + (28.25/18)*5= 15+7.8472 = 22.8472.<BR>\n" );
document.write( "    * Q3 is in 15-19. Therefore, Q3=15 + [(188.25-160)/18]*5 = 15 + 7.8472 = 22.8472\r<BR>\n" );
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document.write( "* **Quartile Deviation (QD):**<BR>\n" );
document.write( "    * QD = (Q3 - Q1) / 2 = (22.8472 - 10.0917) / 2 = 12.7555 / 2 = 6.3778\r<BR>\n" );
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document.write( "**c. Standard Deviation and Coefficient of Variation**\r<BR>\n" );
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document.write( "* **Standard Deviation (σ):**<BR>\n" );
document.write( "    * σ = √[Σ(f(x - μ)²) / Σf]<BR>\n" );
document.write( "    * Calculate Σ(f(x - μ)²) = 1963.6<BR>\n" );
document.write( "    * σ = √(1963.6 / 250) = √7.8544 = 2.8026\r<BR>\n" );
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document.write( "* **Coefficient of Variation (CV):**<BR>\n" );
document.write( "    * CV = (σ / μ) * 100 = (2.8026 / 12.18) * 100 = 23.01%\r<BR>\n" );
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document.write( "**d. Skewness**\r<BR>\n" );
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document.write( "* **Pearson's Coefficient of Skewness:**<BR>\n" );
document.write( "    * Skewness = 3(Mean - Median) / Standard Deviation<BR>\n" );
document.write( "    * Skewness = 3(12.18 - 12.1667) / 2.8026 = 3(0.0133) / 2.8026 = 0.0399 / 2.8026 = 0.0142<BR>\n" );
document.write( "    * Since the skewness is very close to 0, the distribution is nearly symmetrical.\r<BR>\n" );
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document.write( "**e. Percentiles**\r<BR>\n" );
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document.write( "* **60th Percentile (P60):**<BR>\n" );
document.write( "    * P60 position = (60/100) * 250 = 150th observation<BR>\n" );
document.write( "    * P60 is in the 10-14 class.<BR>\n" );
document.write( "    * P60 = 10 + [(150 - 60) / 150] * 5 = 10 + (90 / 150) * 5 = 10 + 3 = 13\r<BR>\n" );
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document.write( "* **90th Percentile (P90):**<BR>\n" );
document.write( "    * P90 position = (90/100) * 250 = 225th observation<BR>\n" );
document.write( "    * P90 is in the 20-24 class.<BR>\n" );
document.write( "    * P90 = 20 + [(225 - 220) / 12] * 5 = 20 + (5 / 12) * 5 = 20 + 2.0833 = 22.0833\r<BR>\n" );
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document.write( "**f. Five-Number Summary (Box Plot)**\r<BR>\n" );
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document.write( "* Minimum: 0<BR>\n" );
document.write( "* Q1: 10.0917<BR>\n" );
document.write( "* Median: 12.1667<BR>\n" );
document.write( "* Q3: 22.8472<BR>\n" );
document.write( "* Maximum: 34<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );