Answer to Question #270425 in Statistics and Probability for Light

Question

The weight of bags of rice sold by a food retailer were classified as follows:

 Weight (kg)

  0 to under 1.5

 1.5 to under 3.0

  3.0 to under 4.5

 4.5 to under 6.0

  6.0 to under 7.5

 7.5 to under 9.0

 9.0 under 10.5

  10.5 to under 12.0

 Number of bags

  10

 25

  30

 48

  35

  22

 20

  10

 a) What percentage of the bags weigh less than 4.5 kg?

b) If a bag is randomly selected from this shop what id the probability that it weighs

9.0kg and 12.0kg?

c) Estimate the following statistics for the weights of the bags in this shop:

i) Modal weight

ii) Median weight

iii) Mean weight

iv) Variance

v) Standard deviation

vi) Coefficient of skewness and explain your result.

Answer 1

a). What percentage of the bags weigh less than 4.5 kg?

• Total number of bags = 10+25+30+48+35+22+20+10 = 200
• Number of bags weigh less than 4.5 kg = 10+25+30 = 65

Percentage of the bags weigh less than 4.5 kg = 65/200 = 0.325*100 = 32.5%

b). If a bag is randomly selected from this shop what would be the probability that it weighs 9.0kg and 12.0kg?

• Total number of bags = 10+25+30+48+35+22+20+10 = 200
• Number of bags weighs 9kg and 12 kg = 20+10 = 30

Probability that bags weighs 9kg and 12 kg = 30/200 = 0.15

c.1). Modal weight

The class with the highest frequency is the modal class. Modal class is 4.5-<6

Modal Weight

$M= L + \frac{F_1-F_0}{2F_1-F_0-F_2}\\ M= L + \frac{48-30}{2*48-30-35}=5.371$

c.2). Median Weight

Median Class = n/2th observation = 100th observation. Median class = 4.5-<6

The class with the highest frequency is the modal class. Modal class is 4.5-<6

Median Weight

$M= L + \frac{N/2-CF}{F}*W\\ M= L + \frac{200/2-65}{48}*1.5=5.5938$

c.3). Mean Weight

Mean Weight 

$\bar{X}= \frac{\sum XF}{\sum F}= \frac{1153.5}{200}=5.768$

c.4) Variance

$Variance\\ s^2= \frac{\sum X^2F-\sum (XF)^2/N}{N-1}= \frac{8136-1153.5^2/200}{200-1}=7.453\\ Standard \space deviation\\ s= \sqrt{7.453}=2.73\\ Coefficient \space of \space skewness\\ = \frac{3(5.768-5.5938)}{2.73}=0.1914$