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Length of metal strips produced by a machine process are normally distributed with a mean length of 250cm and a standard devi

Between 200cm and 290cm : 1/3 vered 3/4 wered arks vered Submit all parts 4 marks Try another question like this one Reveal a

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Answer #1

Solution :

Given that ,

mean = \mu = 250

standard deviation = \sigma =20

P(x < 214 ) = P[(x - \mu ) / \sigma < (214-220) /20 ]

= P(z < -0.30 )

= 0.0013

probability =0.0013

b)

P(x > 294) = 1 - p( x< 294 )

=1- p [(x - \mu ) / \sigma < (294-220) /20 ]

=1- P(z < 3.7)

= 1 - 0.9999 = 0.0001

probability = 0.0001

c)

P( 200< x < 290) = P[(200-220)/20 ) < (x - \mu ) /\sigma  < (290-220) /20 ) ]

= P(-1 < z < 3.5 )

= P(z < 3.5) - P(z < -1 )

Using standard normal table

= 0.9998- 0.1587 = 0.8411

Probability = 0.8411

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