Question

1. in order to estimate the normal snowfall (in inches) for january for all u.s. cities, a random sample of 19 u.s. cities yielded the following data rounded to the nearest inch.

source: http://lwf.ncdc.noaa.gov/oa/climate/online/ccd/snowfall.html
0 1 1 1 1 2 3 5 7 7 7 7 9 9 12 20 21 47 66
a) construct a stem-and-leaf diagram of the data set. make sure you include a key. (4 pts)
b) describe the distribution of the data set. (4 pts)
c) calculate the following statistics for the data. (1 pt. each)
a. mean =
b. median=
c. q₁=
d. q₃=
e. standard deviation =
f. variance =
g. iqr =
h. range =
d) identify any potential outliers using the upper and lower limits. you must show work. (4 pts)
e) determine the appropriate measure of center and variation for this data set. explain why these meas preferable to other measures of center and variation. (4 pts)
center:
variation:
explanation:
f) obtain the z-score for the snow fall of 66 inches. based on the z-score, is 66 inches an unusual v your answer. (4 pts)

Explanation:

Step1: Recall the formula for the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step2: Identify the values of \(x\), \(\mu\), and \(\sigma\)

From part c), we have \(x = 66\), \(\mu=11.89\), and \(\sigma = 16.54\).

Step3: Substitute the values into the formula

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Step4: Determine if the value is unusual

A value is considered unusual if \(|z|> 2\). Since \(z = 3.27>2\), 66 inches is an unusual value.

Answer:

The z - score for 66 inches is approximately \(3.27\). Since \(|z|=3.27>2\), 66 inches is an unusual value.