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LenKa [72]
3 years ago
5

We have IQ test scores of 31 seventh-grade girls in a Midwest school district. We have calculated that sample mean is 105.84 and

the standard deviation is 14.27.
Give a 99% confidence interval for the average score in the population. What is the margin of error?
Mathematics
1 answer:
yuradex [85]3 years ago
4 0

Answer:

The 99% confidence interval for the average score in the population is between 99.24 and 112.44.

The margin of error is 6.60.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.575*\frac{14.27}{\sqrt{31}} = 6.60

The lower end of the interval is the mean subtracted by M. So it is 105.84 - 6.60 = 99.24

The upper end of the interval is the mean added to M. So it is 105.84 + 6.60 = 112.44

The 99% confidence interval for the average score in the population is between 99.24 and 112.44.

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3 years ago
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zhenek [66]
Hey there Smarty!

This would be considered to be a (acute angel) which in this case, we would have to make sure that this whole triangle would equal less than 270 because each angle would be less than 90°

If we add/multiply \boxed{(90+90+90) \ or \ (90*3) = 270}.

So, we would have to know that was ever this would all add up to be, this would have to be less than 270°

\left[\begin{array}{ccc}\boxed{\boxed{(2.5+5) \\ \\ (2.5*2.5) \\ (2.5-2) \\ (2.5-1) \\}} \\ \\ this \ would \ be \ why \ I \ would \ say \\ that \ this \ would \ be \ the \ answer \end{array}\right]

I truly hope this helps, and also, it's kind of my 
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~Jurgen
7 0
4 years ago
How many five-digit numbers are there that contain the digit 1 (at least) three times in a row?
My name is Ann [436]

Answer:

So you say you want some 5 digit numbers eh? Let’s consider some boundaries. 5 digit numbers range from 10,000 - 99,999. So our number is guaranteed to be between 0 and 90000.

Sometimes it’s much easier to work with concrete things. Such as the number of numbers that don’t contain any 3s. It’s really more difficult to put our finger on “at least 1” anything so we try to steer clear of it.

Anyways we’ve got 90000 numbers that have at least 1 or 0 3’s. Find one and we know the other.

So no 3’s at all right? Well you have 8 choices for the first digit (can’t choose 0 and you can’t choose 3). And then for the remaining 4 digits you have 9 choices. So that’s

8∗94=52488  

Back to our initial equation, we have

90000−8∗94=37512  5 digits numbers without any 3s.

A quick (or maybe not so quick) matlab script confirms such calculations.

credits to : Trevor Squires

7 0
3 years ago
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