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maria ran an experiment to determine the optimal conditions for growing…

Question

maria ran an experiment to determine the optimal conditions for growing artichokes. the data below displays the weights of the artichokes she grew (measured in grams).
45.8 74.1 55.4 62.2 53.4 74.1 49 59 74.1 36.6
what is the mean weight of the artichokes?
mean = \boxed{} grams (round to two decimal places)
what is the median weight of the artichokes?
median = \boxed{} grams
what is the mode weight of the artichokes?
mode = \boxed{} grams

Explanation:

Mean Calculation

Step1: Sum all the weights

The weights are 45.8, 74.1, 55.4, 62.2, 53.4, 74.1, 49, 59, 74.1, 36.6. Let's sum them up:
\(45.8 + 74.1 + 55.4 + 62.2 + 53.4 + 74.1 + 49 + 59 + 74.1 + 36.6\)
First, add step by step:
\(45.8 + 74.1 = 119.9\)
\(119.9 + 55.4 = 175.3\)
\(175.3 + 62.2 = 237.5\)
\(237.5 + 53.4 = 290.9\)
\(290.9 + 74.1 = 365\)
\(365 + 49 = 414\)
\(414 + 59 = 473\)
\(473 + 74.1 = 547.1\)
\(547.1 + 36.6 = 583.7\)

Step2: Divide by the number of data points

There are 10 data points. So the mean is \(\frac{583.7}{10} = 58.37\) (rounded to two decimal places).

Median Calculation

Step1: Order the data

First, order the weights from smallest to largest:
36.6, 45.8, 49, 53.4, 55.4, 59, 62.2, 74.1, 74.1, 74.1

Step2: Find the middle value(s)

Since there are 10 data points (even number), the median is the average of the 5th and 6th values.
The 5th value is 55.4 and the 6th value is 59.
So the median is \(\frac{55.4 + 59}{2} = \frac{114.4}{2} = 57.2\)

Mode Calculation

Step1: Identify the most frequent value

Looking at the ordered data: 36.6, 45.8, 49, 53.4, 55.4, 59, 62.2, 74.1, 74.1, 74.1
The value 74.1 appears 3 times, which is more than any other value. So the mode is 74.1.

Answer:

s:
Mean: \(\boxed{58.37}\) grams
Median: \(\boxed{57.2}\) grams
Mode: \(\boxed{74.1}\) grams