Question

(U.3413) 4. A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 20 cm and variance 2.56 cm. Calculate: a) the probability that a component is at least 24 cm long; (0.0062) b) the probability that the length of a component lies between 19 and 21 cm (0.4681)

question no4 in detailed

Answer 1

Solution trom w given Un cualho Mean (M) = 20 cm Variana 12) = 2-56 cm t = 52.56 [O=16 ray the probability that a component is at least 24 cm long pl$t>24) ?? plaszy)= P19 AL 24-M) "P(Z> 24.7.2005 =P(Z3 4 16) = P(Z > 2.5) . Plissu | P( x S290 - 0 - 60 62 ) (b) P[ 19 2x221) = ? P(1920221) = P(19-M LX M 2 21 - M.) =P(19.20 L Z 2 2.20 ) ap( i zzato) 0-625 < ZL 0.625) EP1720.625) - P172-0.625) - 0.73401 - 0 2659 PC 192 XL21) 0.4681 si so, the required probability sy ) 0:46844,