A computer supply house receives a large shipment of blank CDs each week. Past experience has shown that the number of flaws per CD can be described by the following probability distribution, where the random variable X represents the number of flaws per CD.
x
P(X = x)
0
0.65
1
0.20
2
0.10
3
0.05
(A) Calculate the mean and standard deviation of the number of flaws per CD.
(B) Suppose you select two CDs at random. What is the sampling distribution of the average number of flaws on the two CDs? That is, what is the sampling distribution of the sample mean X ?
(C) Suppose you select 100 CDs at random. What is the (approximate) sampling distribution of the sample mean X? Why?
(D) Suppose you select 100 CDs at random. Approximate the probability that the sample mean Xis less than 0.4.
(A) To calculate the mean and standard deviation of the number of flaws per CD, we multiply each value of x by its probability and sum the results:
Mean (μ) = Σ (x * P(X = x))
= 0 * 0.65 + 1 * 0.20 + 2 * 0.10 + 3 * 0.05
= 0 + 0.20 + 0.20 + 0.15
= 0.55
Next, we can calculate the standard deviation (σ) using the formula:
Standard Deviation (σ) = √[Σ (x - μ)² * P(X = x)]
First, we calculate the difference between each value of x and the mean (x - μ), square it, multiply by the probability, and sum the results:
[(0 - 0.55)² * 0.65] + [(1 - 0.55)² * 0.20] + [(2 - 0.55)² * 0.10] + [(3 - 0.55)² * 0.05]
= [(-0.55)² * 0.65] + [(0.45)² * 0.20] + [(1.45)² * 0.10] + [(2.45)² * 0.05]
= [0.3025 * 0.65] + [0.2025 * 0.20] + [2.1025 * 0.10] + [6.0025 * 0.05]
= 0.196625 + 0.0405 + 0.21025 + 0.300125
= 0.7475
Finally, we take the square root of the above result to obtain the standard deviation:
Standard Deviation (σ) = √0.7475 ≈ 0.865
Therefore, the mean number of flaws per CD is 0.55, and the standard deviation is approximately 0.865.
(B) The sampling distribution of the average number of flaws on the two CDs (sample mean) can be approximated using the Central Limit Theorem. According to this theorem, for a large enough sample size, the sampling distribution of the sample mean will approach a normal distribution.
Therefore, the sampling distribution of X can be approximated as a normal distribution with the same mean as the population mean (0.55) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (2 CDs).
Sampling Distribution (X) ≈ N(0.55, σ/sqrt(n))
(C) If we select 100 CDs at random, the sampling distribution of the sample mean X will still be approximately normally distributed due to the Central Limit Theorem. The mean of the sampling distribution will remain the same (0.55). However, the standard deviation of the sampling distribution will decrease as the sample size increases.
Sampling Distribution (X) ≈ N(0.55, σ/sqrt(n))
(D) To approximate the probability that the sample mean X is less than 0.4 for a sample size of 100 CDs, we can use the sampling distribution discussed in part (C). We standardize the sample mean using the formula:
Z = (X - μ) / (σ / sqrt(n))
Where X is the value we want to find the probability for (0.4), μ is the population mean (0.55), σ is the population standard deviation (0.865), and n is the sample size (100).
Z = (0.4 - 0.55) / (0.865 / sqrt(100))
= -0.15 / (0.865 / 10)
= -0.15 / 0.0865
= -1.734
Now, we need to find the probability associated with Z = -1.734 using a standard normal distribution table or calculator. The approximate probability that the sample mean X is less than 0.4 can be obtained as:
P(X < 0.4) ≈ P(Z < -1.734)
Using a standard normal distribution table or calculator, the probability P(Z < -1.734) is approximately 0.0416.
Therefore, the approximate probability that the sample mean X is less than 0.4 for a sample size of 100 CDs is 0.0416.
(A) To calculate the mean and standard deviation of the number of flaws per CD, we can use the following formulas:
Mean (μ) = ∑(x * P(X = x))
Standard Deviation (σ) = √[∑((x-μ)^2 * P(X = x))]
Using the values given in the probability distribution, we can calculate the mean:
Mean = (0 * 0.65) + (1 * 0.20) + (2 * 0.10) + (3 * 0.05)
= 0 + 0.20 + 0.20 + 0.15
= 0.55
To calculate the standard deviation, we need to first calculate the squared deviations from the mean:
(0-0.55)^2 * 0.65 = 0.3575
(1-0.55)^2 * 0.20 = 0.055
(2-0.55)^2 * 0.10 = 0.2405
(3-0.55)^2 * 0.05 = 0.09025
Adding these values and taking the square root gives us the standard deviation:
Standard Deviation = √(0.3575 + 0.055 + 0.2405 + 0.09025)
= √0.74325
≈ 0.8622
Therefore, the mean number of flaws per CD is 0.55 and the standard deviation is approximately 0.8622.
(B) If we select two CDs at random, the sampling distribution of the average number of flaws on the two CDs follows a normal distribution (central limit theorem). The mean of the sampling distribution of the sample mean (X-bar) will be the same as the population mean (0.55), and the standard deviation of the sampling distribution (σ-bar) can be obtained by dividing the population standard deviation (0.8622) by the square root of the sample size (2):
σ-bar = σ / √n
= 0.8622 / √2
≈ 0.6104
Therefore, the sampling distribution of the sample mean X-bar, when selecting two CDs at random, follows a normal distribution with a mean of 0.55 and a standard deviation of approximately 0.6104.
(C) If we select 100 CDs at random, the sampling distribution of the sample mean X will also follow a normal distribution (central limit theorem). The mean of the sampling distribution of the sample mean (X-bar) will again be the same as the population mean (0.55), but the standard deviation (σ-bar) will be obtained by dividing the population standard deviation (0.8622) by the square root of the sample size (100):
σ-bar = σ / √n
= 0.8622 / √100
= 0.8622 / 10
= 0.08622
Therefore, the sampling distribution of the sample mean X-bar, when selecting 100 CDs at random, follows a normal distribution with a mean of 0.55 and a standard deviation of approximately 0.08622.
The sampling distribution of the sample mean becomes approximately normal regardless of the shape of the original distribution due to the central limit theorem. As the sample size increases, the sampling distribution becomes more and more normal.
(D) To approximate the probability that the sample mean X is less than 0.4 when selecting 100 CDs at random, we need to calculate the z-score corresponding to 0.4:
z = (0.4 - μ) / σ
= (0.4 - 0.55) / 0.08622
≈ -1.738
Using a standard normal distribution table or calculator, we can find that the probability of a z-score less than -1.738 is approximately 0.0416.
Therefore, the approximate probability that the sample mean X is less than 0.4 when selecting 100 CDs at random is approximately 0.0416.