Question

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	4	5	6	7	8	9	10	11
Frequency	7	8	9	9	5	5	6	8

Round all your answers to 4 decimal places where possible.

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?

Answer 1

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