رياضيات الإحصاء - دراسة معطيات

# إحصاء رياضي ## Chapter 2 ### Measures of central tendency * **Measures of location** 1. **The Arithmetic mean (A.M)**: * الوسط الحسابي * $A.M = X = \frac{\sum_{i=1}^{n}x_i}{n} = \frac{x_1 + x_2 +...+x_n}{n}$ * **Example 1:** The ages of ten students are:...

# إحصاء رياضي ## Chapter 2 ### Measures of central tendency * **Measures of location** 1. **The Arithmetic mean (A.M)**: * الوسط الحسابي * $A.M = X = \frac{\sum_{i=1}^{n}x_i}{n} = \frac{x_1 + x_2 +...+x_n}{n}$ * **Example 1:** The ages of ten students are: 16, 20, 19, 21, 18, 20, 17, 22, 20, 17. Determine the mean age. * **Solution:** $X = \frac{16 + 20 + 19 + 21 + 18 + 20 + 17 + 22 + 20 + 17 }{10}$ $=\frac{190}{10} = 19$. * Let the frequency of the number *x*₁ be *f*₁, the frequency of the number *x*₂ be *f*₂, and so on. Then, $X = \frac{\sum_{i=1}^{n}f_i x_i}{ \sum_{i=1}^{n}f_i}$ (1) * **Example 2:** Find the mean for the following data: 2 5 6 5 2 6 8 5 8 6 6 7 9 * **Solution:** $X = \frac{2 + 5 + 6 + 5 + 2 + 6 + 8 + 5 + 8 + 6 + 6 + 7 + 9}{13}$ $= \frac{(2*2)+(5*3) + (6×4)+(8*2) + 7 + 9}{2 + 3 + 4 +2+1+1}$ $= \frac{75}{13}$ $= 5.77$. * Note: Equation (1) is the same one we use for the mean of grouped data (data that is tabulated). ### 2 **The Harmonic mean (H.M)** * The reciprocal of the arithmetic mean of the reciprocals of the numbers. * $H.M = \frac {n}{ \sum_{i=1}^{n} \frac{1}{x_i} } = \frac{\sum_{i=1}^{n} f_i}{ \sum_{i=1}^{n} \frac{f_i}{x_i} }$ * **Example 4:** Find the H.M for 2, 4, 8, 11, 4 * **Solution:** $H.M = \frac{5}{ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{11} + \frac{1}{4} } = \frac{5}{(107/80)}$ $= 4.112$. * **Notes:** 1. Extreme values have the least effect. 2. The formula breaks down when "Zero" is one of the observations. ### 3 **The Geometric mean (G.M)**: * $G.M = \sqrt[n]{x_1 . x_2 ... x_n} $ * $= \sqrt[n]{(x_1^{f_1})*(x_2^{f_2})* ... *(x_n^{f_n})}$ * **Example 5:** Find the G.M for 5, 8, 12, 25. * **Solution** $G.M = \sqrt[4]{5*8*12*25} = 10.466$. * **Notes:** 1. The formula breaks down when "Zero" is one of the observations. 2. In general, H.M <= G.M <= A.M . The equality signs hold only if all the observations are identical. 3. $ln G.M = ln(\sqrt[n]{x_1 . x_2 ... x_n} )= \frac{1}{n} \sum_{i=1}^{n} ln x_i $ 4. $= \frac{ln x_1 + ln x_2 + ... + ln x_n}{n}$ 5. $= \frac{ \sum_{i=1}^{n} f_i ln x_i + \sum_{i=1}^{n} f_i ln x_2 + ... + \sum_{i=1}^{n} f_i ln x_n }{ \sum_{i=1}^{n} f_i}$ 6. $= \frac{ \sum_{i=1}^{n} f_i ln x_i} { \sum_{i=1}^{n} f_i} $ * **Example:** For the following frequency table: | Interval | Frequency | |---|---| | 3-7 | 8 | | 7-11 | 2 | | 11-15 | 13 | | 15-19 | 5 | | 19-23 | 3 | | 23-27 | 7 | | 27-31 | 2 | Find: * Arithmetic mean * Harmonic mean * Geometric mean * **Solution** * $A.M = \frac{608}{40} = 15.2$ * $H.M = \frac{40}{3.6} = 11.1$ * $ln G.M = \frac{103.27}{40} = 2.58$ * $G.M = e^{2.58} = 13.2$ ### 4 **The median:** الوسيط * The value of the variable that divides a distribution into two equal parts when the values are arranged in order of magnitude. * $X = \begin{cases} \frac{x_n + 1}{2} \quad if \quad n \quad odd \\ \frac{x_n}{2} \quad + \quad \frac{x_{n+1}}{2} \quad if \quad n \quad even \end{cases}$ * **Example 6:** Find the median for: 1. 15, 22, 17, 25, 15, 19, 21 2. 4, 5, 6, 8, 9, 10, 12, 15 * **Solution:** * 1. n = 7 odd $\implies$ $X = x_4 = \frac{x_3+ x_4}{2}$ * $x_1 \quad x_2 \quad x_3 \quad x_4 \quad x_5 \quad x_6 \quad x_7$ * $15 \quad 15 \quad 17 \quad 19 \quad 21 \quad 22 \quad 25$ * $\implies X = 19$. * 2. n = 8 even $\implies$ $X = \frac{x_4+x_5}{2} = \frac{8+9}{2} = 8.5$. ### **Median for grouped data:** * $X= L+ (\frac{\frac{n}{2} - f_c }{f_{median}} ) * W$ * **L** $\implies$ Lower class boundary of the median class * **n/2** $\implies$ n/2 * **f_c** $\implies$ Cumulative frequency before the median class. * **f_median** $\implies$ Frequency of median class. * **W** $\implies$ Width or class size * **Example 7:** The table below shows the height of 70 men randomly selected at Ismailia. | Height | Frequency | |---|---| | 117-126 | 8 | | 126-135 | 10| | 135-144 | 14 | | 144-153 | 18 | | 153-162 | 9 | | 162-171 | 7 | | 171-180 | 4| Compute the median. * **Solution:** * n/2 = /frac{\sum f_i}{2}= \frac{70}{2} = 35 * 8 + 10 + 14 + 18 = 50 > 35 * Therefore, the median class is 144-153 * $X = 144 + (\frac{35 - 32}{18}) * 9 = 144+ 1.5$ * $X = 145.5$. * **Note:** The median is not affected by extreme values. * **Ordered Data:** H... Q1 ... median... Q3... D5... P99 * **Q:** quartile الربيع * **D:** decile عشر * **P:** percentile مئيات * *Q*₁ = *L* + ( /frac{n}{4} - *f_c* / *f_Q*) * *W* * *D*₁ = *L* + ( /frac{n}{10} - *f_c* / *f_D*) * *W* * *P*₁ = *L* + ( /frac{n}{100} - *f_c* / *f_P*) * *W* ### 5 **The mode:** المنوال * Is the value of the data which occurs most frequently. * **EX:** 1. 2, 5 2, 7, 3, 8 have mode = 2 (unimodal) 2. 2, 5, 2, 7, 5, 8 have mode = 2 and 5 (bimodal) 3. 2, 5, 2, 7, 5, 7, 8, 11 have mode = 2, 5 and 7. (multimodal) 4. 2, 5, 7, 3, 8, 9 have no mode. 5. 2, 5, 2, 3, 5, 7, 8, 5 have mode = 5. ### **Mode for grouped data:** * Mode = *L* + (*\frac{Δ_1}{Δ_1 + Δ_2}*) * *W* * **L** $\implies$ Lower class boundary of the modal class. * **Δ_1** $\implies$ The difference between the frequency of the modal class and the previous class. * **Δ_2** $\implies$ The difference between the frequency of the modal class and the next class. * **W** $\implies$ Width of the class. * **Example 8:** For the table below, find the mode. | Class | Frequency | |---|---| | 10-20 | 6 | | 20-30 | 20 | | 30-40 | 12 | | 40-50 | 10 | | 50-60 | 9 | | 60-70 | 9 | * **Solution** * The modal class is 20-30 as it has the highest frequency (20). * Mode = 20 + ( /frac{20-6}{ (20-6) + (20-12) } ) * 10 = 20 + ( /frac{14}{14+8} ) * 10 = 20 + ( /frac{140}{22} ) = 26.4 ### **Notes:** 1. **For symmetric distribution:** Mode = median = mean 2. **For asymmetric distribution:** * Mean - Mode = 3 (Mean - median) * Or * Mode = 3 median - 2 mean * In case of unimodal distributions.