Statistics I Solved Paper 2025 2nd Annual FBISE

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Statistics I Solved Paper 2025 2nd Annual FBISE. In this post, we are going to discuss, Statistics I HSSC I FBISE Solved Paper 2025 2nd Annual, MCQS, Short Questions, Extensive Questions. Dive into the world of statistics with our comprehensive solutions to the Statistics I paper, covering key topics such as Introduction to Statistics, Measures of Central Tendency and Dispersion, Index Numbers, Correlation & Regression, and Time Series Analysis. Perfect for students and professionals alike, our detailed explanations will help you master these fundamental concepts. Explore more resources on statistical, economical, accounting, and finance topics at bcfeducation.com.

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Table of Contents

Statistics I Solved Paper 2025 2nd Annual FBISE

Q.2: Attempt any Fourteen parts.

(i) Differentiate between discrete and continuous variable.

Answer:

Discrete Variable

A variable which can assume only some specific values within a given range is called discrete variable. For e.g. Number of students in a class, Number of houses in a street, number of children in a family etc. Discrete variable is countable.

Continuous Variable

A variable which can assume any value within a given range is called a continuous variable. For example, age of persons, speed of car, temperature, height, etc. Continuous variable is measurable.

(ii) What methods are used for collection of primary data?

Answer:

The most common methods or sources of collecting primary data are:

  • Questionnaires
  • Surveys
  • Interviews
  • Polls
  • E-mail & Telephone

(iii) Read the following paragraph and make the frequency distribution of number of letters in each word with one as class interval.

“Studying books is the main source of knowledge. Books are indeed never failing friends of man. For a mature mind, reading is the greatest source of pleasure and solace to distressed minds”.

Answer:

Number of Letters (Class Interval = 1)Frequency
11
26
36
42
54
65
73
83
91
101
Total32 words

(iv) A student obtained the following marks in four subjects. Compute weighted arithmetic mean.

SubjectEnglishUrduPhysicsChemistry
Marks (X)50608090
Weight (W)2143

Solution:

SubjectMarks (X)Weight (W)WX
English502100
Urdu60160
Physics804320
Chemistry903270
  ∑W=10∑WX = 750
𝐗𝐰=∑WX∑W=𝟕𝟓𝟎𝟏𝟎=𝟕𝟓{\overline{\mathbf{X}}}_{\mathbf{w}}\mathbf{=}\frac{\sum WX}{\sum W}\mathbf{=}\frac{\mathbf{750}}{\mathbf{10}}\mathbf{= 75}

(v) Calculate arithmetic mean of a frequency distribution of X, if:

U= X−60010, ∑fu= −20, ∑f=5 U = \ \frac{X – 600}{10},\ \sum fu = \ – 20,\ \sum f = 5\

Solution:

𝐀.𝐌 𝐗=𝐀+∑𝐟𝐮∑𝐟×𝐡\mathbf{A.M\ }\overline{\mathbf{X}}\mathbf{= A +}\frac{\mathbf{\sum fu}}{\mathbf{\sum f}}\mathbf{\times h}

𝐀.𝐌 𝐗=𝟔𝟎𝟎+−𝟐𝟎𝟓×𝟏𝟎\mathbf{A.M\ }\overline{\mathbf{X}}\mathbf{= 600 +}\frac{\mathbf{- 20}}{\mathbf{5}}\mathbf{\times 10}

𝐀.𝐌 𝐗=𝟔𝟎𝟎−𝟐𝟎𝟎𝟓=𝟓𝟔𝟎\mathbf{A.M\ }\overline{\mathbf{X}}\mathbf{= 600 -}\frac{\mathbf{200}}{\mathbf{5}}\mathbf{= 560}

(vi) Compute Geometric and Harmonic mean of: 10, 15, 20, 30

Solution:

X1/X
100.1
150.066
200.05
300.033
 ∑ (1/X) = 0.249
𝐆.𝐌=(𝟏𝟎×𝟏𝟓×𝟐𝟎×𝟑𝟎)𝟏/𝟒=𝟏𝟕.𝟑𝟐 \mathbf{G.M =}\left( \mathbf{10 \times 15 \times 20 \times 30} \right)^{\mathbf{1/4}}\mathbf{= 17.32}\

𝐇.𝐌=𝐧∑(𝟏𝐗)=𝟒𝟎.𝟐𝟒𝟗=𝟏𝟔.𝟎𝟔 \mathbf{H.M =}\frac{\mathbf{n}}{\sum_{}^{}\left( \frac{\mathbf{1}}{\mathbf{X}} \right)}\mathbf{=}\frac{\mathbf{4}}{\mathbf{0.249}}\mathbf{= 16.06}\

(vii) The following is a frequency distribution of the number of children born in 65 families. Find median and mode for the data:

No. of Children (X)012345
No. of families (f)471422126

Solution:

XfC.F
044
1711
21425
32247
41259
5665
 ∑f = n 65 
𝐌𝐞𝐝𝐢𝐚𝐧= 𝐧+𝟏𝟐=𝟔𝟓+𝟏𝟐=𝟑𝟑 \mathbf{Median = \ }\frac{\mathbf{n + 1}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{65 + 1}}{\mathbf{2}}\mathbf{= 33}\

33 falls in C.F of 47 so X = 3 is median

Mode = Maximum frequency is 22 so X = 3 is mode

(viii) If lower quartile is 20 and upper quartile is 40. Calculate semi-interquartile range and coefficient of quartile deviation.

Solution:

Quartile Deviation

𝐐𝐃 𝐨𝐫 𝐒.𝐈.𝐐.𝐑=𝐐𝟑−𝐐𝟏𝟐 \mathbf{QD\ or\ S.I.Q.R =}\frac{\mathbf{Q}_{\mathbf{3}}\mathbf{-}\mathbf{Q}_{\mathbf{1}}}{\mathbf{2}}\

𝐐𝐃 𝐨𝐫 𝐒.𝐈.𝐐.𝐑=𝟒𝟎−𝟐𝟎𝟐=𝟏𝟎 \mathbf{QD\ or\ S.I.Q.R =}\frac{\mathbf{40 – 20}}{\mathbf{2}}\mathbf{= 10}\

Coefficient of Quartile Deviation

𝐐𝟑−𝐐𝟏𝐐𝟑+𝐐𝟏=𝟒𝟎−𝟐𝟎𝟒𝟎+𝟐𝟎=𝟐𝟎𝟔𝟎=𝟎.𝟑𝟑 \frac{\mathbf{Q}_{\mathbf{3}}\mathbf{-}\mathbf{Q}_{\mathbf{1}}}{\mathbf{Q}_{\mathbf{3}}\mathbf{+}\mathbf{Q}_{\mathbf{1}}}\mathbf{=}\frac{\mathbf{40 – 20}}{\mathbf{40 + 20}}\mathbf{=}\frac{\mathbf{20}}{\mathbf{60}}\mathbf{= 0.33}\

(ix) A student calculated mean and variance of 25 observations as 20 and 16 respectively. Compute coefficient of variation.

Solution:

Data

n=25, X=20, Var=16, S.D=4 n = 25,\ \overline{X} = 20,\ Var = 16,\ S.D = 4\

C.V=S.DX×100 C.V = \frac{S.D}{\overline{X}} \times 100\

C.V=420×100=20 C.V = \frac{4}{20} \times 100 = 20\

(x) If Median = 20 and Arithmetic mean = 22, compute the value of mode.

Solution:

Mode=3Median-2Mean

Mode=3(20)-2(22)

Mode=60-44=16

(xi) Calculate mean deviation from mean and mean coefficient of dispersion of 6, 8, 10, 12, 14

Solution:

X|X-Mean|
64
82
100
122
144
∑X = 50∑|X-Mean|=12
Mean=∑Xn=505=10 Mean = \frac{\sum X}{n} = \frac{50}{5} = 10\

M.D=∑|X−Mean|n=125=2.4 M.D = \frac{\sum|X – Mean|}{n} = \frac{12}{5} = 2.4\

Coefficient of Mean Deviation

C.M.D=Mean deviationMean C.M.D = \frac{Mean\ deviation}{Mean}\

C.M.D=2.410=0.24 C.M.D = \frac{2.4}{10} = 0.24\

(xii) Compute price index numbers taking “average of first 03 years” as base.

Year2000200120022003200420052006
Price20152528405060

Solution:

Step 1: Find Base Price (Average of First 3 Years)

Prices for 2000, 2001, 2002:

20+15+253=20 \frac{20 + 15 + 25}{3} = 20\

YearPricePrice Index
200020(20/20) 100 = 100
200115(15/20) 100 = 75
200225(25/20) 100 = 125
200328(28/20) 100 = 140
200440(40/20) 100 = 200
200550(50/20) 100 = 250
200660(60/20) 100 = 300

 (xiii) Given that ∑p1qo = 9925, ∑poqo = 9270, ∑p1q1 = 10710, ∑poq1 = 10000. Compute Laspeyres, Paasches and Fisher’s Price Index number.

Solution:

Laspeyre′sIndex=∑p1qo∑poqo× 100 Laspeyre’s Index = \frac{\sum p_{1}q_{o}}{\sum p_{o}q_{o}} \times \ 100\

Laspeyre′sIndex=99259270 × 100 Laspeyre’s Index = \frac{9925}{9270}\ \times \ 100\

Laspeyre′sIndex= 107.06 Laspeyre’s Index = \ 107.06\

Paasche′sIndex=∑p1q1∑poq1× 100 Paasche’s Index = \frac{\sum p_{1}q_{1}}{\sum p_{o}q_{1}} \times \ 100\

Paasche′sIndex=1071010000× 100 Paasche’s Index = \frac{10710}{10000} \times \ 100\

Paasche′sIndex=107.1 Paasche’s Index = 107.1\

Fisher′sIdealIndex=L×P Fisher’s Ideal Index = \sqrt{L \times P}\

Fisher′sIdealIndex=107.06 × 107.1 Fisher’s Ideal Index = \sqrt{107.06\ \times \ 107.1}\

Fisher′sIdealIndex=107.07 Fisher’s Ideal Index = 107.07\

(xiv) Differentiate between simple and composite index number.

Answer:

Simple Index

In simple index number price or quantity of a single product is taken. Simple Index can be calculated through fixed base method and chain base method.

Composite Index

In composite index, the price or quantity is taken related to multiple products. Composite index is further categorized in to unweighted composite index and weighted index.

(xv) If rxy = 0.97 and bxy = 0.82, then calculate byx.

Solution:

rxy=byx×bxy r_{xy} = \sqrt{b_{yx} \times b_{xy}}\

0.97=byx×0.82 0.97 = \sqrt{b_{yx} \times 0.82}\

Taking square on both sides

(0.97)2=byx×0.82 (0.97)^{2} = b_{yx} \times 0.82\

byx=0.94090.82=1.1474 b_{yx} = \frac{0.9409}{0.82} = 1.1474\

(xvi) If Sx = 10, Sy = 8 and rxy = 0.75, then compute two regression coefficients Y on X and X on Y.

Solution:

Regression Coefficient Y on X

byx=rxy(SySx) b_{yx} = r_{xy}\left( \frac{S_{y}}{S_{x}} \right)\

byx=0.75(810)=0.6 b_{yx} = 0.75\left( \frac{8}{10} \right) = 0.6\

Regression Coefficient X on Y

byx=rxy(SxSy) b_{yx} = r_{xy}\left( \frac{S_{x}}{S_{y}} \right)\

byx=0.75(108)=0.9375 b_{yx} = 0.75\left( \frac{10}{8} \right) = 0.9375\

(xvii) Describe the properties of correlation coefficient (r).

Answer:

  • Symmetric Property

Correlation between X variable and Y variable and Y Variable and X variable have same meaning or:

𝐫𝐱𝐲=𝐫𝐲𝐱 \mathbf{r}_{\mathbf{xy}}\mathbf{=}\mathbf{r}_{\mathbf{yx}}\
  • Free from unit of measurement

Correlation coefficient is free from unit of measurement means if data is given in kilograms or length, the answer will not be in kilograms or length.

  • Independent to Origin and Scale

Correlation Coefficient is independent to origin and scale means:

rxy=ruv where U= X±Ah and V=Y±Bh r_{xy} = r_{uv}\ where\ U = \ \frac{X \pm A}{h}\ and\ V = \frac{Y \pm B}{h}\

  • Range of Correlation Coefficient

Correlation Coefficient always lies between -1 to +1 both inclusive.

  • Geometric Mean of Regression Coefficients

Correlation Coefficient is a geometric mean of two regression coefficients.

𝐫𝐱𝐲=𝐛𝐲𝐱 ×𝐛𝐱𝐲 \mathbf{r}_{\mathbf{xy}}\mathbf{=}\sqrt{\mathbf{byx\ \times bxy}}\

  • Relation with Covariance

If X and Y are two independent variables then Cov(X, Y) = 0 but it does not mean that if Cov (X, Y) = 0 then X and Y are definitely independent.

(xviii) If the second degree parabola fitted to the data for the years 1980-87 (both inclusive) with the origin at the middle of 1983 and 1984 is Ŷ = 131.81 + 16.89X + 1.64X², the unit of X being half year, then find the trend values for 1980-87.

Solution:

YearsX= Year – 1983.5X²Ŷ = 131.81 + 16.89XŶ = 131.81 + 16.89X + 1.64X²
1980-3.512.2572.695224.595
1981-2.56.2589.585231.645
1982-1.52.25106.475241.975
1983-0.50.25123.365255.585
19840.50.25140.255272.475
19851.52.25157.145292.645
19862.56.25174.035316.095
19873.512.25190.925342.825
𝟏𝟗𝟖𝟑+𝟏𝟗𝟖𝟒𝟐=𝟏𝟗𝟖𝟑.𝟓 \frac{\mathbf{1983 + 1984}}{\mathbf{2}}\mathbf{= 1983.5}\

(xix) What is meant by irregular or random variation in time series?

Answer:

These are sudden changes occurring in a time series. These are sudden in nature that is why they cannot be explained through trends or seasonal or cyclical movements. For example, sudden change due to tsunami or earthquake etc.

Statistics I Solved Paper 2025 2nd Annual FBISE

Section C- (Marks 26)

Q.3: The following table shows the marks obtained by the students:

Marks10—1920—2930—3940—4950—5960—69
f58101296
  • Calculate the Arithmetic Mean, Median and Mode
  • Compute variance, standard deviation and coefficient of skewness.

Solution:

Income (Rs)Frequency (f)Class BoundariesXfXfX²C.F
10—1959.5—19.514.572.51051.255
20—29819.5—29.524.5196480213
30—391029.5—39.534.534511902.523
40—491239.5—49.544.55342376335
50—59949.5—59.554.5490.526732.2544
60—69659.5—69.564.538724961.550
 50  202593212.5 
 ∑f =  ∑fX =∑fX² = 
  • Calculate the Arithmetic Mean, Median and Mode
X‾=∑fx∑f=202550=40.5\bar{X} = \frac{\sum fx}{\sum f} = \frac{2025}{50} = 40.5

n/2=50/2=25 so model class is 39.5—49.5

X~=L+hf(n2−C)\widetilde{X} = L + \frac{h}{f} \left( \frac{n}{2} – C \right)

X~=39.5+1012(25−23) \tilde{X} = 39.5 + \frac{10}{12} \left( 25 – 23 \right)\

X~=39.5+2012 \tilde{X} = 39.5 + \frac{20}{12}\

X~=41.16 \tilde{X} = 41.16\

X^=L+fm−f1(fm−f1)+(fm−f2)×h \hat{X} = L + \frac{f_m – f_1}{(f_m – f_1) + (f_m – f_2)} \times h\

X^=39.5+12−10(12−10)+(12−9)×10 \hat{X} = 39.5 + \frac{12 – 10}{(12 – 10) + (12 – 9)} \times 10\

X^=39.5+25×10 \hat{X} = 39.5 + \frac{2}{5} \times 10\

X^=39.5+205 \hat{X} = 39.5 + \frac{20}{5}\

X^=43.5 \hat{X} = 43.5\
  • Compute variance, standard deviation and coefficient of skewness.
𝐕𝐚𝐫 (𝐒𝟐)=[∑𝐟𝐗𝟐∑𝐟− (∑𝐟𝐗∑𝐟)𝟐] \mathbf{Var\ (}\mathbf{S}^{\mathbf{2}}\mathbf{) =}\left\lbrack \frac{\mathbf{\sum}\mathbf{f}\mathbf{X}^{\mathbf{2}}}{\mathbf{\sum}\mathbf{f}}\mathbf{-}\mathbf{\ }\left( \frac{\mathbf{\sum}\mathbf{f}\mathbf{X}}{\mathbf{\sum}\mathbf{f}} \right)^{\mathbf{2}} \right\rbrack\

𝐕𝐚𝐫 (𝐒𝟐)=[𝟗𝟑𝟐𝟏𝟐.𝟓𝟓𝟎− (𝟐𝟎𝟐𝟓𝟓𝟎)𝟐] \mathbf{Var\ (}\mathbf{S}^{\mathbf{2}}\mathbf{) =}\left\lbrack \frac{\mathbf{93212.5}}{\mathbf{50}}\mathbf{-}\mathbf{\ }\left( \frac{\mathbf{2025}}{\mathbf{50}} \right)^{\mathbf{2}} \right\rbrack\

𝐕𝐚𝐫 (𝐒𝟐)=[𝟏𝟖𝟔𝟒.𝟐𝟓− 𝟏𝟔𝟒𝟎.𝟐𝟓] \mathbf{Var\ (}\mathbf{S}^{\mathbf{2}}\mathbf{) =}\left\lbrack \mathbf{1864.25}\mathbf{-}\mathbf{\ 1640.25} \right\rbrack\

𝐕𝐚𝐫 (𝐒𝟐)=𝟐𝟐𝟒 \mathbf{Var\ (}\mathbf{S}^{\mathbf{2}}\mathbf{) = 224}\

𝐒.𝐃(𝐒)=[∑𝐟𝐗𝟐∑𝐟− (∑𝐟𝐗∑𝐟)𝟐] \mathbf{S.D(S)}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{f}\mathbf{X}^{\mathbf{2}}}{\mathbf{\sum}\mathbf{f}}\mathbf{-}\mathbf{\ }\left( \frac{\mathbf{\sum}\mathbf{f}\mathbf{X}}{\mathbf{\sum}\mathbf{f}} \right)^{\mathbf{2}} \right\rbrack}\

𝐒.𝐃(𝐒)=[𝟐𝟐𝟒] \mathbf{S.D(S)}\mathbf{=}\sqrt{\left\lbrack \mathbf{224} \right\rbrack}\

𝐒.𝐃(𝐒)=𝟏𝟒.𝟗𝟔 \mathbf{S.D}\left( \mathbf{S} \right)\mathbf{= 14.96}\

𝐂.𝐨𝐟 𝐒𝐤𝐞𝐰𝐧𝐞𝐬𝐬=𝐌𝐞𝐚𝐧−𝐌𝐨𝐝𝐞𝐒.𝐃 \mathbf{C.of\ Skewness =}\frac{\mathbf{Mean – Mode}}{\mathbf{S.D}}\

𝐂.𝐨𝐟 𝐒𝐤𝐞𝐰𝐧𝐞𝐬𝐬=𝟒𝟎.𝟓−𝟒𝟑.𝟓𝟏𝟒.𝟗𝟔 \mathbf{C.of\ Skewness =}\frac{\mathbf{40.5 – 43.5}}{\mathbf{14.96}}\

𝐂.𝐨𝐟 𝐒𝐤𝐞𝐰𝐧𝐞𝐬𝐬=−𝟎.𝟐𝟎 \mathbf{C.of\ Skewness = – 0.20}\

Q.4 (a) Compute index numbers of prices from the following data taking 1970 as base using Arithmetic Mean as an average.

YearAverage Price in Rupees
FirewoodSoft CokeKerosene Oil
19703.252.500.20
19713.442.800.22
19723.502.000.25
19733.752.500.30

Solution

YearsFirewoodP.RSoft CokeP.RKerosene OilP.RSum of Price RelativesArithmetic Mean
19703.251002.51000.2100300300/3 = 100
19713.44105.852.81120.22110327.85327.85/3 = 109.28
19723.5107.692800.25125312.69312.69/3 = 104.29
19733.75115.382.51000.3150365.38365.38/3 = 121.79

(b) Find regression line Y on X and Correlation Coefficient (r): n = 8, ∑X = 36, ∑Y = 152, ∑XY = 768, ∑X²=204, ∑Y²=3056

Solution

Correlation Coefficient

X‾=∑xn=368=4.5 \bar{X} = \frac{\sum x}{n} = \frac{36}{8} = 4.5\

Y‾=∑yn=1528=19 \bar{Y} = \frac{\sum y}{n} = \frac{152}{8} = 19\

σx=[∑x2n− (∑xn)2] \sigma x = \sqrt{\left\lbrack \frac{\sum x^{2}}{n} – \ \left( \frac{\sum x}{n} \right)^{2} \right\rbrack}\

σx=[2048− (4.5)2] \sigma x = \sqrt{\left\lbrack \frac{204}{8} – \ (4.5)^{2} \right\rbrack}\

σx=25.5−20.25 \sigma x = \sqrt{25.5 – 20.25}\

σx=2.29 \sigma x = 2.29\

σy=[∑y2n− (∑yn)2] \sigma y = \sqrt{\left\lbrack \frac{\sum y^{2}}{n} – \ \left( \frac{\sum y}{n} \right)^{2} \right\rbrack}\

σy=[30568− (19)2] \sigma y = \sqrt{\left\lbrack \frac{3056}{8} – \ (19)^{2} \right\rbrack}\

σy=382−361 \sigma y = \sqrt{382 – 361}\

σy=4.58 \sigma y = 4.58\

r=∑xyn−(∑xn)(∑yn)σx.σy r = \frac{\frac{\sum xy}{n} – \left( \frac{\sum x}{n} \right)\left( \frac{\sum y}{n} \right)}{\sigma x.\sigma y}\

r=7688−(4.5)(19)(2.29)(4.58) r = \frac{\frac{768}{8} – (4.5)(19)}{(2.29)(4.58)}\

r=96 –85.510.4882 r = \frac{96\ –85.5}{10.4882}\

r=10.510.4882=1 r = \frac{10.5}{10.4882} = 1\

Regression line Y on X

Y^=a+bx \widehat{Y} = a + bx\

b=r(σyσx) b = r\left( \frac{\sigma y}{\sigma x} \right)\

b=1(4.582.29) b = 1\left( \frac{4.58}{2.29} \right)\

b=2 b = 2\

a=Y‾−bX‾ a = \bar{Y} – b\bar{X}\

a=19−2(4.5)=10 a = 19 – 2(4.5) = 10\

Y^=a+bx \widehat{Y} = a + bx\

Y^=10+2x \widehat{Y} = 10 + 2x\

Q.5 (a) Compute (i) 5-years simple moving average (ii) 4 years centered moving average.

Year19901991199219931994199519961997199819992000
Value2023262923293127383433

Solution (i) 5-years simple moving average

YearValue5 Years M.T5 Years M.A
199020
199123
19922612124.2
19932913026
19942313827.6
19952913927.8
19963114829.6
19972715931.8
19983816332.6
199934
200033

Solution (ii) 4 years centered moving average

YearValues4 Years M.T2 Values M.T4 Years M.A Centered
199020
199123
20+23+26+29=98
19922698+101=199199/8 = 24.88
23+26+29+23=101
199329101+107=208208/8 = 26
26+29+23+29=107
199423107+112=219219/8 = 27.38
29+23+29+31=112
199529112+110=222222/8 = 27.75
23+29+31+27=110
199631110+125=235235/8 = 29.38
29+31+27+38=125
199727125+130=255255/8 = 31.88
31+27+38+34=130
199838130+132=262262/8 = 32.75
27+38+34+33=132
199934
200033

(b) Fit a straight line to the following data and calculate trend values:

Year200120022003200420052006200720082009
Value606580739710593111117

Solution

YearValue (Y)XXYX²Trend Values Ŷ =89+7.06x
200160-4-2401660.76
200265-3-195967.82
200380-2-160474.88
200473-1-73181.94
20059700089
20061051105196.06
20079321864103.12
200811133339110.18
2009117446816117.24
∑Y = 801∑XY = 424∑X²= 60 
Y^=a+bx \widehat{Y} = a + bx\

a=∑Yn=8019=89 a = \frac{\sum Y}{n} = \frac{801}{9} = 89\

b=∑XY∑X2=42460=7.06 b = \frac{\sum XY}{\sum X^{2}} = \frac{424}{60} = 7.06\

Y^=a+bx \widehat{Y} = a + bx\

Y^=89+7.06x \widehat{Y} = 89 + 7.06x\

Statistics I HSSC I FBISE Solved Paper 2023, MCQS, Short Questions, Extensive Questions
Solved Paper Statistics I FBISE 2017

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