The distribution law of the random variable X is given by

1. 
   If each seed planted has a probability of 0.8 of germination, construct the distribution law for the number of sprouts from 3 seeds planted.
   

Since one seed planted may or may not germinate, the number of sprouts out of 3 seeds planted X will be a random variable with a binomial distribution. The values ​​of the random variable X are x1=0; x2=1, x3=2, x4=3, and we find the probabilities of accepting these values ​​using Bernoulli's formula:
Solution:

2. 
   A light bulb factory produced 10,000 light bulbs. The probability that any bulb is broken is p=0.0001. From these bulbs, 4 bulbs were taken at risk. Write the distribution law for the number of defective bulbs.
   

Since the number of bulbs is large and the probability of failure is small, it is convenient to use Poisson's formula. Since n=10000, p=0.0001, =pr=1, k=0, 1, 2, 3, 4, based on Poisson's formula:

3. 
   The distribution law of the random variable X is given by:
   

Find the mathematical expectation, variance, root mean square average of a random variable.
М(Х)=100,1+120,2+200,1+250,2+300,4=1+2,4+2+5+12 =22,4.

М(Х²)=1000,1+1440,2+4000,1+6250,2+900×0,4=10+28,8+40+125+360=563,8.
D(Х)=М(Х²)–[М(Х)]²=563,8–(22,4)²=62,04.

Problems.

1. 
   If the probability of each planted tree turning green is 0.9, make a distribution law of the number of trees planted out of 3 turning green.
   
2. 
   The probability of a shot hitting the target is 0.6. Draw a distribution law for the number of hits from 4 bullets fired.
   
3. 
   A coin is tossed twice. Construct a distribution law for the random variable X, which represents the number of times Tails falls.
   
4. 
   Each ball of cotton was risked if there was a 0.001 probability of wilt disease. Draw the distribution law for the number of cases of wilt disease from 2000 bushels of cotton.
   
5. 
   The distribution laws of random variable X are given. Find its expectation, variance, mean square average.
   

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1. 
   The average weight of the calves kept in the barn is 100 kg, the deviation from the average weight is 0.04. What is the probability that the weight of the bull taken at risk is between 99.5 and 100.5 kg?
   

According to the condition of the problem D(X)=0.04; M(X)=100 and 99.5≤X≤100.5. We subtract M(X) = 100 from this inequality:
–0,5≤Х–М(Х)≤0,5
We get the inequality |X–M(X)|≤0.5 which is as strong as this inequality. Here =0.5. We use the Chebyshev inequality:
, р>0,84

2. 
   2% of non-standard products produced in the workshop. Estimate the probability that 970 to 990 of the 1,000 products are considered standard.
   

We denote the number of standard products by X and find M(X), D(X) and . Standard products. It is 98%. So, p=0.98, non-standard probability q=00.02;
Based on the formulas M(X)=pr and D(X)=prq we find: M(X)=10000.98=980, D(X)=10000.980.02=19.6 .
Subtract M(X) =980 from both sides of the inequality 970≤X≤990, which follows from the condition of the problem: –10≤Х– М(Х) ≤ 10,
|X–M(X)|≤10. From here, taking into account that =10, we find based on the Chebyshev inequality:
, Р>0,804.

3. 
   To determine the average yield of cotton planted on an area of ​​2,000 ha, the yield of cotton on an area of ​​1 m2 per hectare was checked. If it is known that the dispersion of yield per hectare does not exceed 9, find the probability that the average yield in the selected area does not exceed the average yield in the entire area by 0.3 centners.
   

According to the condition of the problem p=2000, D(X)=S = 9, =0.3
If we call the yield per hectare a random quantity (X1 - from the first hectare, X2 - from the second hectare, etc.), then it will be the average yield. If M(X1), M(X2), . . ., M(Xp) is the average productivity per hectare, then based on Chebyshev's theorem

4. 
   According to the information of the technical control department, the percentage of defective parts is 2.5%. Estimate the probability that the percentage of defective parts will not exceed 0.005 from the norm set by the technical control department when 8000 parts are checked for the experiment.
   

We use Bernoulli's theorem to solve this problem. According to the condition of the problem: R=0.025 (2.5% brack details), q=1–r=0.975, =0.005.
Then according to Bernoulli's theorem:

5. 
   The probability of occurrence of event A in each experiment is p=0.6. How many experiments must be performed so that the probability that the relative frequency deviates from the constant probability in each experiment is less than 0.1 in absolute value is greater than 0.98?
   

We use Bernoulli's theorem to solve the problem.
Should be ≥0.98. Here we find n:
≤0, 02 or
So, it is necessary to conduct at least p=1200 experiments.

6. 
   Find the probability that the relative frequency of occurrence of an event in 5,000 independent experiments deviates from the probability r=0.8 of the occurrence of the event in each experiment (where this deviation is expected with probability R>0.94).
   

We also use Bernoulli's formula to solve this problem. We have the following:
, P=0,8; q=1–p=0,2; q=5000

, 0,02