1 answer

The breaking strength of a certain rivet used in a machine engine is normally distributed with...

Question:

The breaking strength of a certain rivet used in a machine engine is normally distributed with mean 5500 psi and standard dev

The breaking strength of a certain rivet used in a machine engine is normally distributed with mean 5500 psi and standard deviation 307 psi. A random sample of 16 rivets is taken. What is the probability that the sample mean falls between 5456.25 psi and 5566.01 psi?

Answers

Let X be the breaking strength of the rivet

X~ Normal ( 5500, 307)

Sample size , n= 16

Let X be the sample mean of the 16 samples

X~ Normal ( 5500, \frac{307}{\sqrt{16}} )

P( 5456.25 < X < 5566.01) = P( \frac{5456.25-5500}{307 /\sqrt{16}} <  \frac{\bar{X}-\mu}{\sigma /\sqrt{n}} <   \frac{5566.01-5500}{307 /\sqrt{16}} )

= P( -0.57< z < 0.86)

= P( z < 0.86) - P(z < -0.57)

= P(z < 0.86) - 1 + P(z < 0.57)

= 0.80511 - 1 + 0.71566

= 0.52077

.

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