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ella [17]
2 years ago
15

The diameters of bolts produced in a machine shop are normally distributed with a mean of 5.755.75 millimeters and a standard de

viation of 0.070.07 millimeters. Find the two diameters that separate the top 6%6% and the bottom 6%6%. These diameters could serve as limits used to identify which bolts should be rejected. Round your answer to the nearest hundredth, if necessary.
Mathematics
1 answer:
svet-max [94.6K]2 years ago
4 0

Let <em>D</em> be the random variable denoting the diameter of this shop's bolts, so that <em>D</em> is normally distributed with <em>µ</em> = 5.75 and <em>σ</em> = 0.07. The top 6% and bottom 6% of bolts have diameters <em>d</em>₁ and <em>d</em>₂ such that

P(<em>d</em>₁ < <em>D</em> < <em>d</em>₂) = P(<em>D</em> < <em>d</em>₂) - P(<em>D</em> < <em>d</em>₁) = 0.94 - 0.06

i.e. <em>d</em>₂ is the 94th percentile and <em>d</em>₁ is the 6th percentile, for which

P(<em>D</em> < <em>d</em>₂) = 0.94

P(<em>D</em> < <em>d</em>₁) = 0.06

Convert <em>D</em> to a random variable <em>Z</em> following the standard normal distribution using

<em>Z</em> = (<em>D</em> - <em>µ</em>) / <em>σ</em>

Then

P(<em>D</em> < <em>d</em>₂) = P((<em>D</em> - 5.75) / 0.07 < (<em>d</em>₂ - 5.75) / 0.07)

0.94 = P(<em>Z</em> < (<em>d</em>₂ - 5.75) / 0.07)

→   (<em>d</em>₂ - 5.75) / 0.07 ≈ 1.55477

→   <em>d</em>₂ ≈ 5.86

P(<em>D</em> < <em>d</em>₁) = P((<em>D</em> - 5.75) / 0.07 < (<em>d</em>₁ - 5.75) / 0.07)

0.06 = P(<em>Z</em> < (<em>d</em>₁ - 5.75) / 0.07)

→   (<em>d</em>₁ - 5.75) / 0.07 ≈ -1.55477

→   <em>d</em>₁ ≈ 5.64

So bolts with a diameter between 5.64 mm and 5.86 mm are acceptable.

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We are given a sample with a mean of 500 and a standard deviation of 100.

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