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1. Let ‌\( X \) be a normally distributed random variable with expected value ‌\( \mu = 3.5 \) and standard deviation ‌\( \sigma = 1.3 \). Use tables for cumulative probabilities in the standard normal distribution to answer the questions.

a) What is the probability ‌\( P(X > 2.7) \)?

b) What is the probability ‌\( P(1.5 < X < 4.0) \)?

2. A machine produces aluminum plates which, according to the product specification, should weigh 100 grams. A small deviation from this weight is accepted, but if the weight deviates by more than ±1 gram from this value, the aluminum plate is considered not to comply with the specification. Assume that the weight of the aluminum plates is normally distributed with an expectation of 100 grams and a standard deviation of 0.6 grams.

a) What is the probability that a randomly selected aluminum plate does not comply with the specification?

b) The plates are packed in boxes of 25. The box weighs 50 grams. What is the expected weight (in grams) of the box of 25 aluminum sheets?

c) What is the variance (in grams²)?

d) What is the probability that the box of 25 aluminum plates weighs less than 2545 grams?

3. For a certain type of machine, the number of machine stops during a working day is Poisson distributed with an expected value of 4. There are 32 such machines in a factory. Let the independent random variables \( X_1, \ldots, X_{32} \) be the respective numbers of machine stops for the 32 machines.

The daily average of the number of machine stops per machine is then \( \bar{X} = \frac{1}{32}\sum_{i=1}^{32} X_i \). From the central limit theorem, we know that \( \bar{X} \) will be approximately normally distributed. What is the expected value and standard deviation of this normal distribution?

Answer :

a) To find the probability P(X > 2.7), we can standardize the value using the standard normal distribution. First, we calculate the z-score for 2.7 using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. Substituting the given values, we get z = (2.7 - 3.5) / 1.3 = -0.6154. Using the cumulative probability table for the standard normal distribution, we can find that P(Z > -0.6154) = 1 - P(Z < -0.6154) = 1 - 0.2676 ≈ 0.7324.

b) To find the probability P(1.5 < X < 4.2), we need to calculate the z-scores for both values. For 1.5, the z-score is (1.5 - 3.5) / 1.3 = -1.5385, and for 4.2, the z-score is (4.2 - 3.5) / 1.3 = 0.5385. Using the cumulative probability table, we find P(-1.5385 < Z < 0.5385) ≈ 0.6207 - 0.2946 = 0.3261.

a) To find the probability that a randomly selected aluminum plate does not comply with the specification, we calculate the probability of the weight being outside the range of 99 to 101 grams. We standardize these values using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. For 99 grams, the z-score is (99 - 100) / 0.6 = -1.67, and for 101 grams, the z-score is (101 - 100) / 0.6 = 1.67. Using the cumulative probability table, we find P(Z < -1.67) + P(Z > 1.67) = 2 * (1 - P(Z < 1.67)) ≈ 2 * (1 - 0.9525) ≈ 0.095.

b) The expected weight of the box of 25 aluminum plates can be calculated by multiplying the expected weight of each plate (100 grams) by the number of plates (25) and adding the weight of the box itself (50 grams). Therefore, the expected weight of the box of 25 aluminum plates is 25 * 100 + 50 = 2550 grams.

c) The variance of the box of 25 aluminum plates can be calculated by squaring the standard deviation of each plate (0.6 grams) and multiplying it by the number of plates (25). Therefore, the variance is (0.6^2) * 25 = 9 grams^2.

d) To find the probability that the box of 25 aluminum plates weighs less than 2545 grams, we standardize the value using the formula z = (x - μ) / σ. For 2545 grams, the z-score is (2545 - 2550) / √(25 * 0.6^2) = -0.8333. Using the cumulative probability table, we find P(Z < -0.8333) ≈ 0.2023.

The expected value of X¯, the average number of machine stops per machine, is equal to the expected value of a single machine stop, which is given as 4. The standard deviation of X¯ can

Learn more about standard deviation here: brainly.com/question/29808998

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