Mathematics, 14.07.2020 14:01 gabriel5575

Fredrick hit 14, 18, 13, 12, 12, 16, 13, 12, 1, and 15 home runs in 10 seasons of play. Which statements are correct? Check all that apply. Fredrick’s data set contains an outlier. The median value is 12 home runs. The mean value is about 12.6 home runs. The median describes Fredrick’s data more accurately than the mean. The mean value stays the same when the outlier is not included in the data set.

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Answer from: lisapcarroll

(a) Yes, Fredrick’s data set contains an outlier.

(b) No, the median value is not 12 home runs.

(d) Yes, the median describes Fredrick’s data more accurately than the mean.

(e) No, the mean value doesn't stay the same when the outlier is not included in the data set.

Step-by-step explanation:

We are given that Fredrick hit 14, 18, 13, 12, 12, 16, 13, 12, 1, and 15 home runs in 10 seasons of play.

Firstly, arranging our data set in ascending order we get;

1, 12, 12, 12, 13, 13, 14, 15, 16, 18.

(a) The statement that Fredrick’s data set contains an outlier is true because in our data set there is one value that stands out from the rest of the data, which is 1.

Hence, the outlier value in the data set is 1.

(b) For calculating median, we have to first observe that the number of observations (n) in the data set is even or odd, i.e;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is odd, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations in Fredrick's data set is even, i.e. n = 10.

SO, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{10}{2})^{th} \text{ obs.}+(\frac{10}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(5)^{th} \text{ obs.}+(6)^{th} \text{ obs.} }{2}$

= $\frac{13+13 }{2}$ = 13 home runs

So, the statement that the median value is 12 home runs is not correct.

Mean = $\frac{1+ 12+ 12+ 12+ 13+ 13+14+ 15+ 16+ 18}{10}$

= $\frac{126}{10}$ = 12.6 home runs

So, the statement that the mean value is about 12.6 home runs is correct.

(d) The statement that the median describes Fredrick’s data more accurately than the mean is correct because even if the outlier is removed from the data set, the median value will remain unchanged but the mean value gets changed.

(e) After removing the outlier, the data set is;

12, 12, 12, 13, 13, 14, 15, 16, 18.

Now, the mean of the data = $\frac{12+12+ 12+ 13+ 13+ 14+ 15+ 16+ 18}{9}$

= $\frac{125}{9}$ = 13.89

So, the statement that the mean value stays the same when the outlier is not included in the data set is incorrect.

Answer from: Quest

you never say the total so we cant find the percentage used for petrol

step-by-step explanation:

Answer from: Quest

First, think of what you know. you know that the number t the end of the equation will need to be subtracted because it is the money he lost from buying the lawn mower. and your first two weeks have negative profit, so that tells you that c and d are wrong. now you can plug any of the weeks into a and b to see which one works. i'll start with plugging week 1 into a. $y = 60(1) - 90 \\ y = 60 - 90 \\ y = - 30$ this isn't right because it says your answer for week one should be -90.so now we need to try the same thing with b. $y = 60(1) - 150 \\ y = 60 - 150 \\ y = - 90$ this one got you the right answer, (sometimes if you only check with one of the weeks multiple will work so you need to check with more of the weeks, but it worked out fine here)this means that b is the correct answer.let me know if you need more clarification. i hope this : )