a company manufactures calculators in batches of 45 and claims that the rate of defects is 4%.find the probability of getting exactly 2 defects in a batch of of 45 if the rate of defects is 4%.if the store receives a batch of 45 calculators and finds that there are 2 defective calculators,do they have any reason to doubt the company's claimed rate of defects.
I don't know how to do this
To find the probability of getting exactly 2 defects in a batch of 45 calculators, we can use the binomial probability formula. The formula is as follows:
P(X = k) = (nCk) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes (defects in this case)
- n is the total number of trials (calculators in a batch)
- k is the number of successes we want (defects in this case)
- p is the probability of success in a single trial (rate of defects)
- (nCk) is the binomial coefficient, calculated as n! / (k! * (n - k)!)
Given that the rate of defects is 4%, we can express it as a decimal form: p = 0.04. Plugging these values into the formula:
P(X = 2) = (45C2) * 0.04^2 * (1 - 0.04)^(45 - 2)
Now let's calculate this value:
Step 1: Calculate (nCk) = 45C2:
(45C2) = 45! / (2! * (45 - 2)!)
= 45! / (2! * 43!)
= (45 * 44) / (2 * 1)
= 990
Step 2: Calculate p^k * (1 - p)^(n - k):
0.04^2 * (1 - 0.04)^(45 - 2)
= 0.0016 * 0.9606^43
≈ 0.0016 * 0.4137
≈ 0.000663
Step 3: Multiply (nCk) and p^k * (1 - p)^(n - k):
(45C2) * 0.04^2 * (1 - 0.04)^(45 - 2)
= 990 * 0.000663
≈ 0.656
Therefore, the probability of getting exactly 2 defects in a batch of 45 calculators, assuming the claimed rate of defects is 4%, is approximately 0.656 or 65.6%.
Now, to evaluate whether the store has any reason to doubt the company's claimed rate of defects, we can consider statistical significance. If the probability of getting 2 defects or fewer (cumulative probability) is significantly lower than the claimed rate of defects (4%), it may raise doubts about the company's claim.
To calculate the cumulative probability, we sum the individual probabilities of getting 0, 1, or 2 defects:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
Let's calculate this:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= (45C0) * 0.04^0 * (1 - 0.04)^(45 - 0) + (45C1) * 0.04^1 * (1 - 0.04)^(45 - 1) + (45C2) * 0.04^2 * (1 - 0.04)^(45 - 2)
By substituting the values into the formula and calculating, we find:
P(X ≤ 2) ≈ 0.134
Since the cumulative probability of obtaining 2 defects or fewer is approximately 0.134 or 13.4%, which is reasonably close to the claimed rate of defects (4%), the store does not have strong grounds to doubt the company's claim. However, it is important to note that this is a statistical analysis, and further analysis and data may be required for a more conclusive assessment.