The following data gives the weight for 15 corn cobs (displayed in ascending ord

Question

The following data gives the weight for 15 corn cobs (displayed in ascending order) which were produced using an organic corn fertilizer: 169, 192, 200, 202, 204, 214. 216, 231, 232, 234, 237, 243, 246, 260, 262 Note that DE1ng 3. 342.0 and ? 1 x2 ith corn cob for i - 1,2, ..., 15 754.216.0, where x 1S the weight of the (a) For this random sample of n -15 observations, obtain the mean, the median, the interquartile range and the standard deviation. (b) Among the four statistics in part 3a, which are measures of central tendency and which are measures of dispersion, c) Are there any outliers in this sample? If so, which values are outliers? (d) We produced a quantile-quantile plot for these data. Is it reasonable to assume that the weight is normally distributed? Explain. Normal Q-Q Plot

Explanation / Answer

a) mean = sum of all the values divided by total number of values = 3342/15 = 222.8

median for n is odd is the middle value in the increasing data set = 8th value = 231

formula of interquartile range is = third quartile - first quartile

let's find them

Q1 = (N+1)/4 }th observation = 16/4 = 4th observation in the increasing data set = 202

Q3 = [3*{(N+1)/2}/4 }]th observation = 3* 16 / 4 = 12th observation = 243

IQR = 243 - 202 = 41

Let's used excel to find standard deviation

First enter the data in the excel column

Used command "=STDEV(range of the data)" and then enter then we get

standard deviation = 26.21123

b) Measure of central tendencies = Mean and Median

Measure of Dispersion = Inter Quartile Range (IQR) and Standard deviation(SD)

c) If there are values outside the (mean - 2*SD , mean + 2*SD) then we call them outliers

Lower limit = mean - 2* SD = 222.8 - 2*26.21123 = 170.3775

Upper limit = mean + 2* SD = 222.8 - 2*26.21123 = 275.2225

only one outlier = 169 because it is not within ( 170.3775, 275.2225)

a) Yes because all the points are very close to the straight line

the point estimate of mean = 222.8

the Point estimate of standard deviation = 26.21123

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