Question

3. Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerator is running. Suppose that 7 of these refrigerator have defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order. Let X be the number among the first 6 examined that have a defective compressor. What is the probability that X exceeds its mean value by more than 1 standard deviation
4. The error involved in making a certain measurement is a continuous rv X with pdf
f(x)={0.09375(4−x 2 )0 −2≤x≤2otherwise f(x)={0.09375(4−x2)−2≤x≤20 otherwise compute P(-1<x<="" span=""> </x

Answer 1

Answer on Question #59020 – Math – Statistics and Probability

Question

3. Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerator is running. Suppose that 7 of these refrigerators have defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order. Let X be the number among the first 6 examined that have a defective compressor. What is the probability that X exceeds its mean value by more than 1 standard deviation?

Solution

It is hypergeometric probability distribution with $N = 12, n = 6, M = 7, N - M = 5$.

$$\mu = n \frac{M}{N} = 6 \frac{7}{12} = 3.5$$ $$\sigma = \sqrt{n \frac{M N - M N - n}{N N - 1}} = \sqrt{6 \frac{7}{12} \frac{5}{12} \frac{12 - 6}{12 - 1}} = 0.89$$ $$\mu + \sigma = 3.5 + 0.89 = 4.39$$

The probability that X exceeds its mean value by more than 1 standard deviation is

$$P(X > \mu + \sigma) = 1 - P(X < \mu + \sigma) = 1 - F(\mu + \sigma).$$

We used Excel function HYPGEOM.DIST:

$$F(4.39) = \text{HYPGEOM.DIST}(4.39, 6, 7, 12, \text{TRUE}) = 0.8788.$$ $$P(X > \mu + \sigma) = 1 - 0.8788 = 0.1212.$$

Answer: 0.1212.

Question

4. The error involved in making a certain measurement is a continuous rv X with pdf

$$f(x) = \{0.09375(4 - x^2) \} - 2 \leq x \leq 2 \text{otherwise}$$ $$f(x) = \begin{cases} 0.09375(4 - x^2), & -2 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases}$$

compute $P(-1 < x < 1)$

Solution

$$P(-1 < x < 1) = \int_{-1}^{1} 0.09375(4 - x^2) = 0.09375\left(4x - \frac{x^3}{3}\right)_{-1}^{1} = 0.09375\left(8 - \frac{2}{3}\right) = 0.6875.$$

Answer: 0.6875.

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