Mathematics, 14.07.2019 19:10 la200564

Georgina likes to read during her summer break. the table below shows the number of pages in each book she read last summer and the number of pages in each book she read this summer. number of pages georgina reads during the summer last summer 140 220 190 180 190 this summer 150 170 150 190 180 what inference can someone draw about the data? the books she read this summer had more pages, on average, than last summer. the number of pages in each book varied more last summer than this summer. last summer, the average book had nearly twice the number of pages than the average book has this summer. this summer, the number of pages in a book varied nearly twice as much as the number of pages in last summer's books

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Answer from: Aethis

The number of pages in each book varied more last summer than this summer.

Step-by-step explanation:

Number of Pages Georgina Reads during the Summer

Last Summer This summer

140 150

220 170

190 150

180 190

190 180

Option 1 :The books she read this summer had more pages, on average, than last summer.

$Average = \frac{\text{sum of all observations }}{\text{total observations}}$

$\text{Average of last summer}= \frac{140+220+190+180+190}{5}$

= $\frac{920}{5}$

= 184

$\text{Average of This summer}= \frac{150+170+150+190+180}{5}$

= $\frac{840}{5}$

= 168

Since the average of pages she read this summer is less than last summer .

Hence Option 1 is wrong .

Option 2: The number of pages in each book varied more last summer than this summer.

To check variation we need to find the standard deviation

$\sigma = \sqrt{\frac{\sum(x-\bar{x})^{2} }{n}}$

N is total no. of observations

$\text{average of last summer }\bar{x}=184$

Standard deviation of last summer :

$\sigma = \sqrt{\frac{([150-168]^2+[170-168]^2+[150-168]^2+[190-168]^2+[180-168]^2)}{5}}$

$\sigma = \sqrt{\frac{1280}{5}}$

$\sigma = 2\sqrt{166}$

$\sigma = 25.76$

Thus the standard deviation of last summer is 25.76

Standard deviation of this summer :

$\sigma = \sqrt{\frac{([140-184]^2+[220-184]^2+[190-184]^2+[180-184]^2+[190-184]^2)}{5}}$

$\sigma = \sqrt{\frac{3320}{5}}$

$\sigma = 16$

Thus the standard deviation of this summer is 16

Thus The number of pages in each book varied more last summer than this summer.

Hence Option 2 is correct

Option 3 : Last summer, the average book had nearly twice the number of pages than the average book has this summer.

Last summer average = 184

This summer average = 168

$184\neq 2\times 168$

$184\neq 336$

Hence Option 3 is wrong .

Option 4: This summer, the number of pages in a book varied nearly twice as much as the number of pages in last summer's books

Standard deviation of last summer is 25.76

Standard deviation of this summer is 16

$16\neq2\times25.67$

$16\neq51.34$

Hence Option 4 is wrong .

Thus Option 2 is right .The number of pages in each book varied more last summer than this summer.

Answer from: Quest

Ithink it should be 55.
Will give brainliestthe graph below shows wilson's science scores versus the number of hours spent d

Answer from: Quest

22.

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