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# sta301 1st assignment solution by ash

if u find any mistake so plzzzzz tell

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plz measurement waly ka ans bta dain plz plz plz plz

Jazak Allah

Ash Dear ummmah

 Class Number class limit Frequency Relative Frequency (r.f) Commulative frequency (c.f) Relative comulative frequency r.f * c.f=r.cf 1 148 ­­- 152 6 6/45 = 0.133 6 0.789 2 153 - 157 11 11/45 = 0.244 17 4.148 3 158 - 162 14 14/45 = 0.311 31 9.641 4 163 - 167 9 9/45 = 0.2 40 8 5 168 - 172 3 3/45 = 0.066 43 2.838 6 173 - 177 2 2/45 = 0.044 45 1.98 Total = 45

y relative cumulative frequency kasy ai?

yar ap ne relative cumulative frequency galat nikali hai.   is k lie formula ye hai:  c.f/total frequency.

yaha total frequency 45 hai. for example: first relative frequency:6/45=0.133   second:17/45=0.377  third:31/45=0.688 and so on.

Question 1:

a)   Following are the heights (in centimeters) of 45 female high school students. Prepare a frequency distribution of the data where the class width is 5, the number of classes is 6 and the smallest observation is 148. Also obtain the relative frequency and relative cumulative frequency.

170, 151, 154, 160, 158, 154, 171, 156, 160, 157, 160, 157, 148, 165, 158, 159, 155, 151, 152, 161, 156, 164, 156, 163, 174, 153, 170, 149, 166, 154, 166, 160, 160, 161, 154, 163, 164, 160, 148, 162, 167, 165, 158, 158, 176.

Ans:

Smallest value(X0) = 148

Largest value (xm) = 176

Range = Xm – X0 = 176 – 148 = 28

Class interval = h = Range/ No of classes

= 28 / 6 = 4.6 ~ 5

 Class Number class limit Frequency Relative Frequency Relative cumulative frequency 1 148 ­­-152 6 6/45 = 0.133 0.133 2 153 – 157 11 11/45 = 0.244 0.133+0.244 = 0.377 3 158 – 162 14 14/45 = 0.311 0.377+0.311=0.688 4 163 – 167 9 9/45 = 0.2 0.688+0.2=0.888 5 168 – 172 3 3/45 = 0.066 0.888+0.066=0.954 6 173 - 177 2 2/45 = 0.044 0.954+0.044=0.998 Total = 45

b)     Which level of measurement (scale of measurement) is suitable for the following data in each example?

i) The lake Saif-ul-Muluk is 10,578 feet above from the sea level.

ii) If customers go to the shop for buying shoes and try different size of shoes. The sizes of shoes are 7B, 10D and 12E.

iii) The numbering on T-shirts of players in the cricket team.

iv) The weight of the person is 68 kg.

Question 2:

a)      Find the median, harmonic mean and 23rd percentile for the following frequency distribution.

 Class Boundaries Frequency CF X 1/x F(1/x) 145 – 150 4 4 147.5 0.0067 0.0268 150 – 155 6 10 152.5 0.0065 0.039 155 – 160 28 38 157.5 0.0063 0.1764 160 – 165 58 96 162.5 0.0061 0.3538 165 – 170 64 160 167.5 0.0059 0.3776 170 – 175 30 190 172.5 0.0057 0.171

Median:

Plz calculate it

Harmonic mean:

23rd Percentile:

b)     Explain the concept of Geometric mean? Also find the geometric mean for the following data.

18, 19, 19, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 23, 24, 27, 30, 36

Ans:

Geometric mean G of a set of n positive values X1, X2…Xn is defined as the positive nth root of their product.

 X logX 18 1.2552 19 1.2787 19 1.2787 19 12.787 19 1.2787 20 1.3010 20 1.3010 20 1.3010 20 1.3010 20 1.3010 21 1.3222 21 1.3222 21 1.3222 21 1.3222 22 1.3424 23 1.3617 24 1.3802 27 1.4313 30 1.4771 36 1.5563 ∑logX =

G = antilog 1.33564 = 21.65

U can also calculat ur result from this calculator on side

http://www.sengpielaudio.com/calculator-geommean.htm

U can also calculate Ur ANTILOG from online calculator on the side

http://ncalculators.com/number-conversion/anti-log-logarithm-calcul...

please pay attention the formula of harmonic mean is not correct because it is used in ungrouped data as we know that in the question we have provided group data so please resolve your question with correct mean to say with grouped formula

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