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netineya [11]
4 years ago
11

The weights (to the nearest pound) of some boxes to be shipped are found to be:. Weight 65 68 69 70 71 72 90 95 frequency 1 2 5

8 7 3 2 2 Their mean weight is 72.97 pounds. What is the standard deviation of these weights? The standard deviation, to the nearest tenth, is a0.
Mathematics
1 answer:
Lena [83]4 years ago
7 0
So the mean is 72.97

We need to subtract the mean from each value and square it.
(65-72.97)^2= 63.5209
(68-72.97)^2=24.7009
(69-72.97)^2=15.7609
(70-72.97)^2=8.8209
(71-72.97)^2= 3.8809
(72-72.97)^2=0.9409
(90-72.97)^2=290.0209
(95-72.97)^2=485.3209

Now we add up the new values ( also consider their frequency) and find their mean.
Add the values
63.5209+(2 •24.7009=49.4018)+(5•15.7609=78.8045)+(8•8.8209=70.5672)+(7•3.8809=27.1663)+(3•0.9409=2.8227)+(2•290.0209=580.0418)+(2•485.3209=970.6418)= 1,842.967
Divide by total numburs to find the mean
1,842.967/ 30=61.43223333

The standar deviation is the square root of the mean so is
Square root of 61.43223333=7.837871735
Round to the nearest tenth
Standard Deviation is 7.8



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The equation is 2\frac{1}{8} - \frac{1}{2} .

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5 0
3 years ago
Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to
RUDIKE [14]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

5 0
3 years ago
What is the measure of ACE in the diagram below?<br><br> A. 46<br> B. 104<br> C. 29<br> D. 58
beks73 [17]

Answer:

C

Step-by-step explanation:

The secant- secant angle ACE is half the difference of the measures of the intercepted arcs, that is

∠ ACE = \frac{1}{2} (AE - BD ) = \frac{1}{2} (104 - 46)° = \frac{1}{2} × 58° = 29° → C

3 0
3 years ago
Read 2 more answers
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