Question

3. You have been provided the following values:
23	23	15	19	14
21	48	32	32	32
59	22	62	29	9

a. Construct a stem-and-leaf plot.
b. How many "stems" are in your plot?
c. Is the data skewed to the left or to the right?
d. What is the mean?
e. What is the median?
f. What is the mode?
g. What is the standard deviation?
h. Calculate 1.s.d. below the mean.
i. Calculate 1 s.d. above the mean.
j. Calculate 2 s.d. below the mean.
k. Calculate 2 s.d. above the mean.
l. Calculate 3 s.d. below the mean.
m. Calculate 3 s.d. above the mean.
n. Are there any outliers in the data?
o. If you answered yes to the question above, which one (s)?

Answer 1

Since the data is not printed properly, assuming the data is as follows:
( if there's any mistake from my side in the data, please do let me know in the comment section so that I can rectify the mistake )

S = { 23, 21, 59, 23, 48, 22, 15, 32, 62, 19, 32, 29, 14, 32, 9 }

a) The stem and leaf plot obtained in R language is as follows :

b) There are 4 stems in the stem and leaf plot as shown in the above image.

All the further required results will be obtained using the Descriptive Statistics function available in the data analysis package of the Excel. This can be obtained as follows :

Step 1 : Enter the data in the excel.

Step 2 : Select the Descriptive Statistics function available in the data analysis package under the Data tab of the Excel.

Step 3 : Enter the data range in the first cell and check box denoting summary statistics.

Step 4 : Click on ok and following results would be obtained :

Data
Mean	29.33333333
Standard Error	4.071581729
Median	23
Mode	32
Standard Deviation	15.76916823
Sample Variance	248.6666667
Kurtosis	0.36377758
Skewness	1.050676276
Range	53
Minimum	9
Maximum	62
Sum	440
Count	15

c) Since Skewness = 1.05068, the data can be terned as positively skewed.

d) Mean = 29.33333

e) Median = 23

f) Mode = 32

g) Standard deviation = sd = 15.7692

h) 1 sd below mean = Mean - 1*sd = 13.5642

i) 1 sd above mean = Mean + 1*sd = 45.1025

j) 2 s.d. below the mean = Mean - 2*sd = -2.2050

k) 2 s.d. above the mean = Mean + 2*sd = 60.8717

l) 3 sd below mean = Mean - 3*sd = -17.9742

m) 3 sd above mean = Mean + 3*sd = 76.6408

n) Outliers in a data can be easily determined using the range ( Q1 - 1.5*IQR , Q3 + 1.5*IQR ) i.e. any value that lies outside of this range would be considered as an outlier, where Q1 is the first quartile, Q3 is the third quartile and IQR is the inter quartile range.
Q1 = QUARTILE.EXC(A1:A15,1) = 19
Q3 = QUARTILE.EXC(A1:A15,1) = 32
IQR = Q3 - Q1 = 13
Thus, ( Q1 - 1.5*IQR , Q3 + 1.5*IQR ) = ( -0.5 , 51.5 )
Since, 62 value lies outside the above mentioned range, thus there is a outlier in the given data.

o) 62 is the only outlier.