Question

# 57 randomly selected students were asked the number of pairs of shoes they have. Let X...

57 randomly selected students were asked the number of pairs of shoes they have. Let X represent the number of pairs of shoes. The results are as follows:

 # of Pairs of Shoes Frequency 4 5 6 7 8 9 10 11 7 8 9 9 5 5 6 8

The mean is:

The median is:

The sample standard deviation is:

The first quartile is:

The third quartile is:

What percent of the respondents have at least 6 pairs of Shoes? %

13% of all respondents have fewer than how many pairs of Shoes?

 x f f*m f*m2 cf 4 7 28 112 7 5 8 40 200 15 6 9 54 324 24 7 9 63 441 33 8 5 40 320 38 9 5 45 405 43 10 6 60 600 44 11 8 88 968 51 total 57 418 3370 mean =x̅=Σf*M/Σf= 7.3333 sample Var s2=(Σfm2-(Σfm2)/n)/(n-1)= 5.4405

the mean is Σxf/Σf =(4*7+5*8+6*9+7*9+8*5+9*5+10*6+11*8)/57 =7.3333

median =middle value =(57+1)/2 th value =29 th value =7

sample standard deviation √5.4405 =2.3325

first quartile =middle value of lower half =(14th +15th value)/2 =5

third quartile is=middle value of upper half =(43th +44th value)/2 =(9+10)/2 =9.5

percent of the respondents have at least 6 pairs of Shoes =(24/57)*100=42.1053%

13% of all respondents have fewer than how many pairs of Shoes =5