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nexus9112 [7]
2 years ago
10

If the inspection division of a county weights and measures department wants to estimate the mean amount of

Mathematics
1 answer:
Amanda [17]2 years ago
7 0

If inspection department wants to estimate the mean amount with 95% confidence level with standard deviation 0.05 then it needed a sample size of 97.

Given 95% confidence level, standard deviation=0.05.

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

We assume that the values follow normal distribution. Normal distribution is a probability that is symmetric about the mean showing the data  near the mean are more frequent in occurence than data far from mean.

We know that margin of error for a confidence interval is given by:

Me=z/2* ST/\sqrt{N}

α=1-0.95=0.05

α/2=0.025

z with α/2=1.96 (using normal distribution table)

Solving for n using formula of margin of error.

n=(z/2ST/Me)^{2}

n=(1.96*0.05)^{2} /(0.01)^{2}

=96.4

By rounding off we will get 97.

Hence the sample size required will be 97.

Learn more about standard deviation at brainly.com/question/475676

#SPJ4

The given question is incomplete and the full question is as under:

If the inspection division of a county weights and measures department wants to estimate the mean amount of soft drink fill in 2 liters bottles to within (0.01 liter with 95% confidence and also assumes that standard deviation is 0.05 liter. What is the sample size needed?

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c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

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\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

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