Ahigh volume manufacture of shafts produces 1000 shafts / day to an advertised specification of 0.250 +/- 0.005”. you are given 30 samples from the day’s production and are asked what % of the shafts are acceptable if the average diameter is 0.251” with a standard deviation of 0.0015”

99.62%

Step-by-step explanation:

given details:

shaft details 0.250 +/- 0.005 I.e. 0.245 and 0.255

and \mu =0.251

\sigma = .0015

now P[0.245<X<0.255]\

= P\frac{0.245 -0.251}{.0015} <\frac{x-\mu}{6}<\frac{0.255 - 0.251}{.0015}

=P [-4 <Z<2.67]

=P[Z<2.67] -P[Z<-4]

=0.9962 -0.00                                              [FROM Z TABLES]

=0.9962

=99.62%

x=31 degrees

y=2

Step-by-step explanation: