Question

A manufacturer of electroluminescent lamps knows that the amount of luminescent ink deposited on one of its products is normally distributed with a mean of 1.2 grams and a standard deviation of 0.03 gram. Any lamp with less than 1.14 grams of luminescent ink fails to meet customers’ specifications. A random sample of 25 lamps is collected and the mass of luminescent ink on each is measured. a. What is the probability that at least one lamp fails to meet specifications? b. What is the probability that five or fewer lamps fail to meet specifications? c. What is the probability that all lamps conform to specifications? d. Why is the joint probability distribution of the 25 lamps not needed to answer the previous questions?

Answer 1

a. 0.4382

b. 0.999983

c. 0.5618

d. See below

Explanation:

let amount of ink be "x"

Mean
`\mu=1.2`

standard deviation
`\sigma=0.03`

number of lamps n = 25

a.

Let's convert to z, which has formula:

`z=(x-\mu)/(\sigma)`

So let's find P(x<1.14)

`P(x<1.14)=P((x-\mu)/(\sigma))=P((1.14-1.2)/(0.03))=P(z<-2)=0.0228` [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:

`nCk=(n!)/((n-k)!k!)p^k q^(n-k)`

So,

we know

n = 25

k = 0

p = 0.0228

q = 1-0.028 = 0.9772

Now we have:

`nCk=(n!)/((n-k)!k!)p^k q^(n-k)\\=(25!)/((25-0)!0!)(0.0228)^(0) (0.9772)^25\\=0.5618`

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:

`P(k\leq 5)=0.999983`

Hence, the probability is 0.999983

c.

Probability that ALL LAMPS conform to specification is when P(k=0).

`P(k=0)=(25!)/((25-0)!0!)(0.0228)^(0) (0.9772)^25\\=0.5618`

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.