The purpose for which these washers are intended allows a maximum tolerance in the diameter of $0.496$ to $0.508$ cm, otherwise the washers are considered defective. Determine the percentage of defective washers produced by the machine, assuming the diameters are normally distributed.

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 08

Year : DEC 2014

Mean $(µ) = 0.502$

Standard deviation $(\sigma ) = 0.005$

Let X denote the diameter of washer

$ (0.496 \lt x \lt 0.508)\\ =p\Bigg[\dfrac {0.496-0.505}{0.005} \lt \dfrac {x-m}{\sigma} \lt \dfrac {0.508-0.502}{0.005}\Bigg]\\ =p[-1.2 \lt z \lt 1.2]\\ =P[0 \lt z \lt 1.2]$

$= 2 \times$ area between $‘z = 0’$ to $‘z = 1.2’$

$$= 2 \times 0.3849$$

$$= 0.7698$$

% of washers with diameter $0.496$ to $0.508$ inches $= 76.98\%$

% of defective washer produced by the machinery

$$= 100 – 76.98 = 23.02\% $$