HELP ME NO RANDOM ANSWERS
2 answers:
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1. The first step here is to arrange the data set's form least to greatest,
Sherelle: 26, 39, 56, 58, 60, 62, 65, 66, 66, 68, 71, 72, 72, 73, 74, 75, 81, 83, 84, 85
Venita: 44, 45, 51, 51, 53, 53, 55, 57, 58, 62, 65, 66, 69, 69, 70, 73, 75, 77, 78, 79
Now we can determine our 5 - number summary based on the numbers respective positions.
First Data Set,
<u>(Five - Number Summary) - Minimum : 26, Quartile 1 : 60, Median : 69.5, Quartile 3 : 75, Maximum : 85</u>
Second Data Set,
<u>(Five - Number Summary) - Minimum : 44, Quartile 1 : 53, Median : 63.5, Quartile 3 : 73, Maximum : 79</u>
2. This part is based on your drawings of the box and whisker plots, so you would have to figure that part out by yourself.
3. First off we know that our data set is composed of the years from 1900, so let's rewrite the set based off of the actual year -
Sherelle: 1926, 1939, 1956, 1958, 1960, 1962, 1965, 1966, 1966, 1968, 1971, 1972, 1972, 1973, 1974, 1975, 1981, 1983, 1984, 1985
Venita: 1944, 1945, 1951, 1951, 1953, 1953, 1955, 1957, 1958, 1962, 1965, 1966, 1969, 1969, 1970, 1973, 1975, 1977, 1978, 1979
( a ) Now in Sherelle's defence, she can say that the lowest coin date in her group is 1926, comparative to Venita's group - the lowest coin date in hers being 1944. Therefore, she is more likely to have the 1916 coin, after all that date is the lowest overall in both their data set.
( b ) In Venita defence, she can say that the mean of her data set is lower than the mean of Sherelle's data set. Take a look at the calculations below,
Sherella's Mean : =
= 1966.8,
Venita's Mean : =
= 1962.5
( c ) I would say Sherella's bag would most likely contain the 1916 coin. The mean is a prominent factor, but their mean(s) only differ by a very small quantity. That too, Sherella's bag contains the lowest coin in both their groups, and though that is not a prominent factor, it could be that she does have the 1916 coin.
