The length of needles produced by a machine has standard deviation of 0.01 inches. assuming that the distribution is normal, how large a sample is needed to determine with a precision of ±0.005 inches the mean length of the produced needles to 95% confidence?

Let X be the length of needle produced by machine. X follows Normal distribution with standard deviation σ=0.01

The margin of error ME = 0.005 and confidence interval is 95%

We have to find here sample size n from given information

The formula to find sample size when population standard deviation is known is

n =

where is critical z value for 95% confidence interval

We have confidence level, c=0.95

α = 1- c = 1-0.95 = 0.05

α/2 = 0.05 /2 = 0.025

z (0.025) is z score value for which probability below -z is 0.025 and above z is 0.025

Using z score table to get z critical value

z = -1.96

For confidence interval calculation we use positive z score value 1.96

The sample size will be then

n =

n = 15.3664

Rounding sample size to nearest integer n=15

a sample is needed to determine with a precision of ±0.005 inches the mean length of the produced needles to 95% confidence is 15

geometry: triangles

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice