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notsponge [240]
3 years ago
6

PLZ HELP ME PLZZZ 15 PTS Mr. Rodriguez asked 60 randomly chosen students from each grade level at Willowbrook School about their

favorite types of reading materials. Of those 60 students, 12 chose science fiction. Mr. Rodriguez used the data to draw the inference that about 20% of middle school students prefer to read science fiction. If 150 students attend the school, did he make a reasonable inference? Explain.
Mathematics
2 answers:
vredina [299]3 years ago
8 0
12/60 students chose science fiction

Approximately x/150 students prefer sf

60x = 1800

x = 30

30/150 students are assumed to prefer sf

30/150 = x/100

150x = 3000

x = 20

20/100 students are likely to prefer sf

Mr. Rodriguez made a reasonable estimate for the approximate percentage of students that prefer science fiction, because if 12/60 is equivalent to 30/150 which refers to the number of students who can be assumed to prefer science Fiction out of the whole school. Considering we need to identify what 30/150 as a percentage is, we can reduce it down to 1/5 to make I easier, then divide 1 by 5 to get .2
.2 as a percentage is 20%, so his inference was indeed reasonable.

(♥ω♥*)Brainliest Please(♥ω♥*)

Artemon [7]3 years ago
3 0

Answer:

Mr. Rodriguez made a reasonable inference. His sampling was random, nonbiased, and reasonably large compared to the school population. 12 is 20% of 60.

Step-by-step explanation:

This is the answer Edge gave me after entering two wrong answers that the previous people on Brainly gave me.

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a) 654.16-2.01\frac{164.55}{\sqrt{52}}=608.29    

654.16+2.01\frac{164.55}{\sqrt{52}}=700.03    

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

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Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that t_{\alpha/2}=2.01

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654.16-2.01\frac{164.55}{\sqrt{52}}=608.29    

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And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

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ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

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n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189

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