Question

Среди выпускаемой цехом продукции 10% является браком. на испытание поступила партия из 10 деталей. каково наивероятнейшее число бракованных деталей в рассматриваемой выборке и какова его вероятность?

Answer 1

1 из 10 будет бракованая
вероятность брака 1/10 тоесть 0,1

Answer 2

Calculation of the Probability of Defective Parts in the Sample

To calculate the probability of the number of defective parts in the sample, we can use the binomial distribution formula. The binomial distribution is used when there are two possible outcomes (defective or non-defective) and each outcome has a fixed probability of occurring.

In this case, we are given that 10% of the production is defective. Therefore, the probability of a part being defective is 0.1, and the probability of a part being non-defective is 0.9.

Let's denote the number of defective parts in the sample as X. We want to find the maximum number of defective parts in the sample, which means we need to find the largest value of X such that the probability of X or fewer defective parts is greater than or equal to 0.5.

To calculate this, we can use the cumulative distribution function (CDF) of the binomial distribution. The CDF gives the probability of getting up to a certain number of successes (defective parts) in a given number of trials (sample size).

The formula for the CDF of the binomial distribution is:

P(X ≤ k) = Σ (n choose i) * p^i * (1-p)^(n-i), for i = 0 to k

Where: - n is the sample size (number of parts in the sample) - k is the maximum number of defective parts in the sample - p is the probability of a part being defective (0.1 in this case)

We want to find the maximum value of k such that P(X ≤ k) ≥ 0.5.

Let's calculate the probability of having 0, 1, 2, ..., up to 10 defective parts in the sample and find the maximum value of k for which the cumulative probability is greater than or equal to 0.5.

Calculation:

Using the binomial distribution formula, we can calculate the probabilities for each value of k:

- P(X ≤ 0) = (10 choose 0) * 0.1^0 * 0.9^10 = 0.9^10 ≈ 0.3487 - P(X ≤ 1) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 ≈ 0.6513 - P(X ≤ 2) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 ≈ 0.8482 - P(X ≤ 3) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 + (10 choose 3) * 0.1^3 * 0.9^7 ≈ 0.9372 - P(X ≤ 4) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 + (10 choose 3) * 0.1^3 * 0.9^7 + (10 choose 4) * 0.1^4 * 0.9^6 ≈ 0.9808 - P(X ≤ 5) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 + (10 choose 3) * 0.1^3 * 0.9^7 + (10 choose 4) * 0.1^4 * 0.9^6 + (10 choose 5) * 0.1^5 * 0.9^5 ≈ 0.9963 - P(X ≤ 6) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 + (10 choose 3) * 0.1^3 * 0.9^7 + (10 choose 4) * 0.1^4 * 0.9^6 + (10 choose 5) * 0.1^5 * 0.9^5 + (10 choose 6) * 0.1^6 * 0.9^4 ≈ 0.9992 - P(X ≤ 7) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 + (10 choose 3) * 0.1^3 * 0.9^7 + (10 choose 4) * 0.1^4 * 0.9^6 + (10 choose 5) * 0.1^5 * 0.9^5 + (10 choose 6) * 0.1^6 * 0.9^4 + (10 choose 7) * 0.1^7 * 0.9^3 ≈ 0.9998 - P(X ≤ 8) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 + (10 choose 3) * 0.1^3 * 0.9^7 + (10 choose 4) * 0.1^4 * 0.9^6 + (10 choose 5) * 0.1^5 * 0.9^5 + (10 choose 6) * 0.1^6 * 0.9^4 + (10 choose 7) * 0.1^7 * 0.9^3 + (10 choose 8) * 0.1^8 * 0.9^2 ≈ 0.9999 - P(X ≤ 9) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 + (10 choose 3) * 0.1^3 * 0.9^7 + (10 choose 4) * 0.1^4 * 0.9^6 + (10 choose 5) * 0.1^5 * 0.9^5 + (10 choose 6) * 0.1^6 * 0.9^4 + (10 choose 7) * 0.1^7 * 0.9^3 + (10 choose 8) * 0.1^8 * 0.9^2 + (10 choose 9) * 0.1^9 * 0.9^1 ≈ 1.0000 - P(X ≤ 10) = (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 + (10 choose 3) * 0.1^3 * 0.9^7 + (10 choose 4) * 0.1^4 * 0.9^6 + (10 choose 5) * 0.1^5 * 0.9^5 + (10 choose 6) * 0.1^6 * 0.9^4 + (10 choose 7) * 0.1^7 * 0.9^3 + (10 choose 8) * 0.1^8 * 0.9^2 + (10 choose 9) * 0.1^9 * 0.9^1 + (10 choose 10) * 0.1^10 * 0.9^0 ≈ 1.0000

From the calculations, we can see that the maximum number of defective parts in the sample is 1, and the probability of having 1 or fewer defective parts in the sample is approximately 0.6513.

Therefore, the maximum number of defective parts in the sample is 1, and the probability of having 1 or fewer defective parts in the sample is approximately 0.6513

Answer 3

Ответ: Среди выпускаемой цехом продукции 10% является браком. На испытание поступила партия из 10 деталей. Каково наивероятнейшее число бракованных деталей в рассматриваемой выборке и какова его вероятность?

Решение: Эта задача сводится к нахождению моды биномиального распределения с параметрами n = 10 и p = 0.1. Мода биномиального распределения определяется по формуле ⌊ ( n + 1 ) p ⌋ {\\displaystyle \\lfloor (n+1)\\,p\\rfloor }. Подставляя данные значения, получаем:

⌊ ( 10 + 1 ) ⋅ 0.1 ⌋ = ⌊ 1.1 ⌋ = 1 {\\displaystyle \\lfloor (10+1)\\cdot 0.1\\rfloor =\\lfloor 1.1\\rfloor =1}

Таким образом, наивероятнейшее число бракованных деталей в выборке равно 1. Вероятность этого события можно найти по формуле биномиального распределения:

P ( X = 1 ) = ( 10 1 ) 0.1 1 ⋅ 0.9 9 ≈ 0.387 {\\displaystyle P(X=1)={\\binom {10} {1}}0.1^ {1}\\cdot 0.9^ {9}\\approx 0.387}

Ответ: Наивероятнейшее число бракованных деталей в выборке равно 1, его вероятность примерно равна 0.387.

: Биномиальное распределение - Wikiwand