Answer:
a. 0.4382
b. 0.999983
c. 0.5618
d. See below
Step-by-step explanation:
let amount of ink be "x"
Mean [tex]\mu=1.2[/tex]
standard deviation [tex]\sigma=0.03[/tex]
number of lamps n = 25
a.
Let's convert to z, which has formula:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
So let's find P(x<1.14)
[tex]P(x<1.14)=P(\frac{x-\mu}{\sigma})=P(\frac{1.14-1.2}{0.03})=P(z<-2)=0.0228[/tex] [z value from z table]
Now,
For 25 lamps, the probability follows binomial distribution with probabilit of success = 0.0228 and number of success = k
We want P(k≥1) = 1 - P(k=0)
Let's use binomial probability distribution formula:
[tex]nCk=\frac{n!}{(n-k)!k!}p^k q^{n-k}[/tex]
So,
we know
n = 25
k = 0
p = 0.0228
q = 1-0.028 = 0.9772
Now we have:
[tex]nCk=\frac{n!}{(n-k)!k!}p^k q^{n-k}\\=\frac{25!}{(25-0)!0!}(0.0228)^{0} (0.9772)^25\\=0.5618[/tex]
So, The probability is 1 - 0.5618= 0.4382
b.
Now, probability of 5 lamps or fewer is basically
P(k=0) + P(k=1) + P(k=2) + P(k=3) + P(k=4) + P(k=5)
Putting all of these individually into the binomial distribution formula above, we get:
[tex]P(k\leq 5)=0.999983[/tex]
Hence, the probability is 0.999983
c.
Probability that ALL LAMPS conform to specification is when P(k=0).
[tex]P(k=0)=\frac{25!}{(25-0)!0!}(0.0228)^{0} (0.9772)^25\\=0.5618[/tex]
Hence, 0.5618
d. joint probability distribution is needed when there are dependent variables. Since, amount of ink in each lamp is an INDEPENDENT variable, joint probability distribution was not needed.
