Use the change of base rule to find the logarithm to four decimal places.
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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=
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=
=96.4
By rounding off we will get 97.
Hence the sample size required will be 97.
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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?
De Moivre's Theorem states that if a complex number is written in the polar coordinate form [ r (cosθ + 
sinθ)] and you raise it to the power n, then this can be evaluated by raising the modulus (r) to the power and multiply the argument (θ) by the power. This therefore would give r ⁿ [cos (nθ) + sin (nθ)].
let A = ∛ <span>(8 cos (4π / 5) + 8 i sin (4π / 5))
</span>⇒ A = ∛ <span>(8 [cos (4π / 5) + i sin (4π / 5)])
</span>
Now by applying De Moivre's Theorem,
⇒ A =
[cos (
× 
) + sin ( × )
⇒ A = 2 [ cos (
) + sin ()
⇒ A = 2 [0.0117 + 0.01297 ] rads
let A = ∛ <span>(8 cos (4π / 5) + 8 i sin (4π / 5))
</span>⇒ A = ∛ <span>(8 [cos (4π / 5) + i sin (4π / 5)])
</span>
Now by applying De Moivre's Theorem,
⇒ A =
⇒ A = 2 [ cos (
⇒ A = 2 [0.0117 +