Not sure if my rule will be correct, but the equation i would use would be y=27x for the spacers and y=5x for the beads.

for 18 inch: 486 spacers and 90 beads

for 24 inch: 648 spacers and 120 beads

4 0

3 years ago

Bill Connors, a quality control manager at a Menlo Park Electronics Company, knows his company has been making surge protectors

Andrews [41]

Answer:

a) Binomial distribution, with n=20 and p=0.10.

b) P(x>1) = 0.6082

c) P(3≤X≤5) = 0.3118

d) E(X) = 2

e) σ=1.34

Step-by-step explanation:

a) As we have a constant "defective" rate for each unit, and we take a random sample of fixed size, the appropiate distribution to model this variable X is the binomial distribution.

The parameters of the binomial distribution for X are n=20 and p=0.10.

$X\sim B(0.10,20)$

b) The probability of k defective surge protectors is calculated as:

$P(x=k) = \binom{n}{k} p^{k}q^{n-k}$

In this case, we want to know the probability that more than one unit is defective: P(x>1). This can be calculated as:

$P(x>1)=1-(P(0)+P(1))\\\\\\P(x=0) = \binom{20}{0} p^{0}q^{20}=1*1*0.1216=0.1216\\\\P(x=1) = \binom{20}{1} p^{1}q^{19}=20*0.1*0.1351=0.2702\\\\\\ P(x>1)=1-(0.1216+0.2702)=1-0.3918=0.6082$

c) We have to calculate the probability that the number of defective surge protectors is between three and five: P(3≤X≤5).

$P(3\leq X\leq 5)=P(3)+P(4)+P(5)\\\\\\P(x=3) = \binom{20}{3} p^{3}q^{17}=1140*0.001*0.1668=0.1901\\\\P(x=4) = \binom{20}{4} p^{4}q^{16}=4845*0.0001*0.1853=0.0898\\\\P(x=5) = \binom{20}{5} p^{5}q^{15}=15504*0*0.2059=0.0319\\\\\\P(3\leq X\leq 5)=P(3)+P(4)+P(5)=0.1901+0.0898+0.0319=0.3118$