# Statistics

Suppose that you are designing an instrument panel for a large industrial machine. The machine requires the person using it to reach 2 feet from a particular position. The reach from this position for adult women is known to have a mean of 2.8 feet with a standard deviation of .5. The reach for adult men is known to have a mean of 3.1 feet with a standard deviation of .6. Both women’s and men’s reach from this position is normally distributed. If this design is implemented:

What percentage of women will not be able to work on this instrument panel?
What percentage of men will not be able to work on this instrument panel?

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1. for women, 2 ft is 1.6 std below the mean
for men, 2 ft is 3.33 std below the mean.

So, look those up in your Z table to see the break points.

Or, play around at

http://davidmlane.com/hyperstat/z_table.html

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2. .For women:
Mean = 2.8
SD = 0.5

P( x < 2) =
μ = 2.8
σ = 0.5
standardize x to z = (x - μ) / σ
P(x < 2) = P( z < (2-2.8) / 0.5)
= P(z < -1.6) = 0.0548 ---- 5.48 percent of women won't be able to work on this instrument panel
(From Normal probability table)

Men:
Mean = 3.1
SD = 0.6
P( x < 2) =
μ = 3.1
σ = 0.6
standardize x to z = (x - μ) / σ
P(x < 2) = P( z < (2-3.1) / 0.6)
= P(z < -1.8333) = 0.0336 --- 3.36 percent of men won't be able to work on this instrument panel
(From Normal probability table)

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