Question

The diameter X [cm] of certain bolts has the density f(x) = k(x - 0.9)(1.1 - x) for 0.9 < x < 1.1 and 0 for other x. If a defective bolt is one that deviates from 1.00 cm by more than 0.06 cm, what percentage of defectives are expected? for what choice of the maximum possible deviation from 1.00 cm shall we obtain 10% defectives

Answer 1

Here as it is given

f(x) = k (x - 0.9)(1.1- x) ; 0.9 < x < 1.1

so here first we have to find the value of k

so here as we know the property of probability distirbution

=

= k [x² -x³/3 - 0.99x]^1.1_0.9

= k* (1/750)

so that is equal to 1

so,

k = 750

f(x) = 750 (x-0.9) (1.1 - x)

so now we know that  a defective bolt is one that deviates from 1.00 cm by more than 0.06 cm.

so here a non - defecitve bolt has diameter between 0.94 cm and 1.06 cm

so here

Pr(Non defective bolt) =

= 750 [x² -x³/3 - 0.99x]^1.06_0.94

= 99/125 = 0.792

Now,

Pr(Defective pieces) =1 - 0.792 = 0.208

Now we have to defince a value of k that is the deviation from 1.00 cm that will limit the defective percent to 10%.

so,

Pr(Defective) = 0.10

Pr(Non defective) = 1 - 0.10 = 0.90

so, writing it mathematically

= 0.90

750 [x² -x³/3 - 0.99x]^1+k_1-k = 0.90

750 [(1+ k)² -(1+k)³/3 - 0.99 (1 + k) - (1-k)² -(1-k)³/3 - 0.99(1 - k)] = 0.90

15k - 500k³ = 0.90

500k³ - 15k + 0.90 = 0

now it is cubic equation , we have to solve it by trial and error.

so we get it k = 0.073

so here the maximum possible deviation from 1.00 cm is 0.73 cm so we obtain 10% defectives.