Question

What is the probability that you select a vial that is within the acceptable range( in...

What is the probability that you select a vial that is within the acceptable range( in other words you do not detect that the machine shifted)

You randomly select 15 vials what is the probability that you select at least one vial that is within acceptable range?

(Acceptable range 9.65 and 9.95) (Mean 9.96) (sd 0.05)

I do have the answer (0.9988)

I don't understand what formula is being used.

How do I solve this and what formula is being used to solve it?

Homework Answers

Answer #1

We need to find: P( 9.65 < X < 9.95 )

mean = 9.96

sd = 0.05

n = 15

P( 9.65 < X < 9.95 ) = P( ( 9.65 - mean )/(sd/sqrt(n)) < ( X - mean )/(sd/sqrt(n)) < ( 9.95 - mean )/(sd/sqrt(n)) )

P( 9.65 < X < 9.95 ) = P( ( 9.65 - 9.96 )/(0.05/sqrt(15)) < ( X -9.96 )/(0.05/sqrt(15)) < ( 9.95 - 9.96 )/(sd/sqrt(15)) )

P( 9.65 < X < 9.95 ) = P( -0.31/(0.05/sqrt(15)) < t < -0.01/(0.05/sqrt(15)) )

P( 9.65 < X < 9.95 ) = P( -1.6 < t < -0.052 )

P( 9.65 < X < 9.95 ) = P( t < -0.052 ) - P( t < -1.6 )

df = 15 - 1 = 14

P( 9.65 < X < 9.95 ) = 0.479632 -  .065958

P( 9.65 < X < 9.95 ) = 0.4137

As per my knowledge this is the correct answer

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