Question

The following data were obtained from a salary survey of Canadian laboratory employees in 2006. The salaries were rounded to the nearest $10,000 and there were 333 respondents.

See picture.

(a) Determine the mean, median, and mode salaries of these lab employees.

(b) Determine the standard deviation of the average salaries.

(c) Calculate the skew and kurtosis of this distribution (you can assume that the population mean and

standard deviation are the same as the sample values).

3. The folowing data were obtained from a salary survey of Canadian laboratory employees in 2006. The salaries were rounded to the nearest $10,000 and there were 333 respondents. Average Number of Average Number of Salary (S Respondents 110,000 20,000 130,000 140,000 150,000 Average Number of Salary ( Respondents Salary (S) Respondents 0,000 20,000 30,000 40,000 21 15 60,000 70,000 80,000 90,000 100,000 49 52 9 21 39 39 27

Answer 1

Given:

Average Salary ($) x	Number of Respondents f
10,000	2
20,000	5
30,000	9
40,000	21
50,000	39
60,000	49
70,000	52
80,000	33
90,000	39
100,000	27
110,000	21
120,000	15
130,000	9
140,000	6
150,000	6

(a) Determine Mean, Median and Mode:

Mean:

$\dpi{100} Mean,\;\bar{x}=\frac{\sum{fx}}{\sum{f}}$

Average Salary ($) x	Number of Respondents f	fx
10000	2	20000
20000	5	100000
30000	9	270000
40000	21	840000
50000	39	1950000
60000	49	2940000
70000	52	3640000
80000	33	2640000
90000	39	3510000
100000	27	2700000
110000	21	2310000
120000	15	1800000
130000	9	1170000
140000	6	840000
150000	6	900000
	$\dpi{100} \sum{f}=333$	$\dpi{100} \sum{fx}=25630000$

$\dpi{100} Mean,\;\bar{x}=\frac{25630000}{333}=76966.96697\sim \$76,966.97$

The mean salary is $76,966.97

Median:

Mid value of the data is Median. To find the mid value cumulative frequency is calculated as below

Average Salary ($)	Number of Respondents	Cumulative Frequency
10000	2	2
20000	5	7
30000	9	16
40000	21	37
50000	39	76
60000	49	125
70000	52	177
80000	33	210
90000	39	249
100000	27	276
110000	21	297
120000	15	312
130000	9	321
140000	6	327
150000	6	333

Total of Frequency, n = 333

Therefore, the median value would be calculated as

$\dpi{100} Median=\left(\frac{n+1}{2} \right )^{th} value$

Therefore,

$\dpi{100} Median=\left(\frac{333+1}{2} \right )^{th} value=\left(\frac{334}{2} \right )^{th} value=167^{th}value$

The cumulative frequencies just greater than 167 is 177 and the value of x corresponding to 177 is 70,000. Hence, the median salary is $70,000

Mode:

Mode is defined as the most frequently occurred value. Therefore, the salary with highest number of respondents will be the mode. The salary with highest respondent is $70,000 with 52 respondents. Therefore, Mode is $70,000

(b) Determine Standard Deviation:

$\dpi{100} Standard\;Deviation,s=\sqrt{\frac{\sum{fd^2}}{\sum{f}-1}}$

$\dpi{100} d=x-\bar{x}$

Therefore,

Average Salary ($) x	Number of Respondents f	d x - 76966.97	d2	fd2
10000	2	-66966.97	4484575070.9809	8969150141.9618
20000	5	-56966.97	3245235670.9809	16226178354.9045
30000	9	-46966.97	2205896270.9809	19853066438.8281
40000	21	-36966.97	1366556870.9809	28697694290.5989
50000	39	-26966.97	727217470.9809	28361481368.2551
60000	49	-16966.97	287878070.9809	14106025478.0641
70000	52	-6966.97	48538670.9809	2524010891.0068
80000	33	3033.03	9199270.9809	303575942.3697
90000	39	13033.03	169859870.9809	6624534968.2551
100000	27	23033.03	530520470.9809	14324052716.4843
110000	21	33033.03	1091181070.9809	22914802490.5989
120000	15	43033.03	1851841670.9809	27777625064.7135
130000	9	53033.03	2812502270.9809	25312520438.8281
140000	6	63033.03	3973162870.9809	23838977225.8854
150000	6	73033.03	5333823470.9809	32002940825.8854
	$\dpi{100} \sum{f}=333$			$\dpi{100} \sum{fd^2}=271836636636.64$

$\dpi{100} s=\sqrt{\frac{271836636636.64}{333-1}}=\sqrt{\frac{271836636636.64}{332}}=\sqrt{818785050.1104}$

$\dpi{100} s=28614.4203$

Sample standard deviation of the salary is $28,614.42

(c) Calculate Skew and Kurtosis

Skewness and Kurtosis are calculated based on moments

Skew:

$\dpi{100} Skew,\;\beta_1=\frac{\mu_3^2}{\mu_2^3}$

$\dpi{100} \mu_2=\frac{\sum{fd^2}}{\sum{f}}$

$\dpi{100} \mu_3=\frac{\sum{fd^3}}{\sum{f}}$

Average Salary ($) x	Number of Respondents f	d	d2	d3	fd2	fd3
10000	2	-66966.97	4484575070.9809	-300318404241126.0000	8969150141.9618	-600636808482252.0000
20000	5	-56966.97	3245235670.9809	-184871243111699.0000	16226178354.9045	-924356215558494.0000
30000	9	-46966.97	2205896270.9809	-103604263982272.0000	19853066438.8281	-932438375840446.0000
40000	21	-36966.97	1366556870.9809	-50517466852844.8000	28697694290.5989	-1060866803909740.0000
50000	39	-26966.97	727217470.9809	-19610851723417.8000	28361481368.2551	-764823217213294.0000
60000	49	-16966.97	287878070.9809	-4884418593990.8000	14106025478.0641	-239336511105549.0000
70000	52	-6966.97	48538670.9809	-338167464563.8010	2524010891.0068	-17584708157317.7000
80000	33	3033.03	9199270.9809	27901664863.1991	303575942.3697	920754940485.5700
90000	39	13033.03	169859870.9809	2213788794290.2000	6624534968.2551	86337762977317.7000
100000	27	23033.03	530520470.9809	12219493923717.2000	14324052716.4843	329926335940364.0000
110000	21	33033.03	1091181070.9809	36045017053144.2000	22914802490.5989	756945358116028.0000
120000	15	43033.03	1851841670.9809	79690358182571.2000	27777625064.7135	1195355372738570.0000
130000	9	53033.03	2812502270.9809	149155517311998.0000	25312520438.8281	1342399655807980.0000
140000	6	63033.03	3973162870.9809	250440494441425.0000	23838977225.8854	1502642966648550.0000
150000	6	73033.03	5333823470.9809	389545289570852.0000	32002940825.8854	2337271737425110.0000
	$\dpi{100} \sum{f}=333$				$\dpi{100} \sum{fd^2}=271836636636.64$	$\dpi{100} \sum{fd^3}=3011757304327320$

$\dpi{100} \mu_2=\frac{271836636636.64}{333}=816326236.1461$

$\dpi{100} \mu_3=\frac{3011757304327320}{333}=9044316229211.16$

$\dpi{100} \beta_1=\frac{9044316229211.16^2}{816326236.1461^3}=0.1504$

Skew = 0.1504

Kurtosis:

$\dpi{100} Kurtosis,\;\beta_2=\frac{\mu_4}{\mu_2^2}$

$\dpi{100} \mu_4=\frac{\sum{fd^4}}{\sum{f}}$

Average Salary ($) x	Number of Respondents f	d	d4	fd4
10000	2	-66966.97	20111413567263300000.0000	40222827134526700000.0000
20000	5	-56966.97	10531554560206900000.0000	52657772801034300000.0000
30000	9	-46966.97	4865978358327440000.0000	43793805224947000000.0000
40000	21	-36966.97	1867477681625110000.0000	39217031314127300000.0000
50000	39	-26966.97	528845250099856000.0000	20624964753894400000.0000
60000	49	-16966.97	82873783751684100.0000	4060815403832520000.0000
70000	52	-6966.97	2356002580592070.0000	122512134190787000.0000
80000	33	3033.03	84626586580028.7000	2792677357140950.0000
90000	39	13033.03	28852375769648000.0000	1125242655016270000.0000
100000	27	23033.03	281451970129796000.0000	7599203193504490000.0000
110000	21	33033.03	1190676129667020000.0000	25004198723007500000.0000
120000	15	43033.03	3429317574381330000.0000	51439763615720000000.0000
130000	9	53033.03	7910169024272720000.0000	71191521218454500000.0000
140000	6	63033.03	15786023199341200000.0000	94716139196047100000.0000
150000	6	73033.03	28449672819586700000.0000	170698036917520000000.0000
	$\dpi{100} \sum{f}=333$			$\dpi{100} \sum{fd^4}=622476626963180000000$

$\dpi{100} \beta_2=\frac{622476626963180000000}{816326236.1461^2}=2.8051$

Kurtosis = 2.8051