On a production line making cameras, the probability of a randomly chosen camera being substandard is 0.072. Find the probability that there are fewer than 18 cameras which are substandard.
p=0.072(p is the probability of success)
n=300 ( n is the no. of times an exp. is performed)
q=1-0.072 ;q=0.928 (q is the probability of failure)
mean=μ=np ; μ=300x0.072 ; μ=21.6
σ^2=npq ; = 300x0.072x0.928 ; σ^2=20.0448
σ=4.477
p(x<18)=p(x<17.5) (continuity error is applied which is 0.5)
now we will standardize it
p(x<17.5-21.6/4.477) ; p(x<-0.9157)
1- (standard area of the graph 0.9157)
=1-8199
=0.180
http://davidmlane.com/hyperstat/z_table.html
BUT out of how many cameras made? 18 out of what = what?
To find the probability that there are fewer than 18 substandard cameras, we can use the binomial distribution formula. The binomial distribution is applicable here because each camera is treated as a separate trial, and each trial can only have two outcomes: substandard or not substandard.
Let's use the formula:
P(X < 18) = ∑ from k = 0 to 17 (n C k) * p^k * (1 - p)^(n - k)
Where:
P(X < 18) is the probability that there are fewer than 18 substandard cameras,
n is the total number of cameras,
k is the number of substandard cameras,
p is the probability of a camera being substandard.
In this case, the total number of cameras is not given, so we will assume n to be a large number, such as 1000, to approximate the probability accurately.
Let's calculate it step-by-step:
Step 1: Calculate the term (n C k) for each value of k from 0 to 17.
(n C k) = n! / (k! * (n - k)!)
For k = 0:
(1000 C 0) = 1000! / (0! * (1000 - 0)!)
(1000 C 0) = 1
For k = 1:
(1000 C 1) = 1000! / (1! * (1000 - 1)!)
(1000 C 1) = 1000
Similarly, calculate (n C k) for k = 2 to 17.
Step 2: Calculate the term p^k * (1 - p)^(n - k) for each value of k from 0 to 17.
For k = 0:
p^0 * (1 - p)^(1000 - 0) = 1 * (1 - 0.072)^(1000 - 0) = 0.928^1000
For k = 1:
p^1 * (1 - p)^(1000 - 1) = 0.072 * (1 - 0.072)^(1000 - 1) = 0.072 * 0.928^999
Similarly, calculate p^k * (1 - p)^(n - k) for k = 2 to 17.
Step 3: Calculate the sum of all the terms.
P(X < 18) = (n C 0) * p^0 * (1 - p)^(n - 0) + (n C 1) * p^1 * (1 - p)^(n - 1) + ... + (n C 17) * p^17 * (1 - p)^(n - 17)
Substitute the values calculated in steps 1 and 2 to find the final probability.
To find the probability that there are fewer than 18 cameras that are substandard, we need to use the binomial distribution formula.
The binomial distribution is used when there are only two possible outcomes for each trial (in this case, a camera being substandard or not substandard), and the probability of success (a camera being substandard) remains the same for each trial. The number of trials is fixed (the total number of cameras in this case), and each trial is independent.
Let's denote the probability of success (a camera being substandard) as p = 0.072, and the number of trials as n, which represents the total number of cameras. We want to calculate the probability of having fewer than 18 substandard cameras, which can be represented as P(X < 18).
The formula for the binomial distribution is:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of having exactly k substandard cameras.
C(n, k) is the combination of n choose k, which represents the number of ways to choose k substandard cameras out of n total cameras.
p^k is the probability of k substandard cameras.
(1 - p)^(n - k) is the probability of the remaining (n - k) cameras being not substandard.
Now, to find the probability of having fewer than 18 substandard cameras, we need to calculate the probabilities for all possible values less than 18 (from 0 to 17) and sum them up.
P(X < 18) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 17)
We can use a calculator or statistical software to calculate each individual probability. Alternatively, we can use a binomial distribution table to look up the values if available.
Once we have calculated or looked up the individual probabilities, we can sum them up to get the final answer.