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Statistics 8th Edition Jia Junping Chapter 5 probability and probability distribution
2022-07-06 14:29:00 【No two or three things】
One 、 Examination site induction
Two 、 Exercises 1 Write the sample space for the following randomized trials :
(1) Record the average score of a statistical test in a class ;
(2) Someone is riding a bike on the road , Observe the number of times the cyclist meets the green light before stopping at the first red light ;
(3) Produce products until 10 Until it is genuine , Record the total number of products produced .
Explain :(1) The average score is in the range 0~100 A continuous variable between , So the sample space of the average score Ω=[0,100].
(2) The number of green lights encountered is from 0 Any natural number at the beginning , So sample space Ω=N.
(3) There may be no defective products or any number of defective products in the products previously produced , So sample space Ω={10,11,12,13,…}.
2 Someone's flower 2 Yuan to buy lottery , He hit 100 The probability of a yuan prize is 0.1%, win a lottery 10 The probability of a yuan prize is 1%, win a lottery 1 The probability of a yuan prize is 20%, Suppose all kinds of prizes cannot be won at the same time , Try to ask for :
(1) The probability distribution of this person's income .
(2) The expected value of this person's income .
Explain :(1) set up X For the benefit of this person , win a lottery 100 The income of yuan prize is 100-2=98( element ), win a lottery 10 The income of yuan prize is 10-2=8( element ), win a lottery 1 When you win a prize, you will lose 1 element , win a lottery 0 When you win a prize, you will lose 2 element , Then the probability distribution of this person's income is :
P(X=98)=0.001
P(X=8)=0.01
P(X=-1)=0.2
P(X=-2)=1-P(X=98)-P(X=8)-P(X=-1)=0.789
(2)E(X)=98×0.001+8×0.01+(-1)×0.2+(-2)×0.789=1.6
3 Set the random variable X The probability density of is :
f(x)=3x2/θ3,0<x<θ
(1) It is known that P(X>1)=7/8, seek θ Value .
(2) seek X Expectation and variance of .
Explain :(1) from P(X>1)=7/8 Available :P(X≤1)=1-P(X>1)=1/8, namely
Solution :θ=2. therefore , A random variable X The probability density of is :f(x)=3x2/8,0<x<2.
(2) from X The expected value of the probability density is :
also
therefore
4 There is 5 questions , At the same time, each question lists 4 An alternative answer , One of the answers is correct . A student can answer correctly by guessing at least 4 What's the probability of the problem ?
Explain : set up X Is the number of correct answers . From the meaning of the question , The probability of answering a question correctly is 1/4. that X~B(5,1/4), So the answer is right, at least 4 The probability of the problem is :
5 Set the random variable X The compliance parameter is λ The Poisson distribution of , And known P{X=1}=P{X=2}, seek P{X=4}. Explain : The formula of Poisson distribution has :
P(X=1)=λe-λ
P(X=2)=λ2e-λ/(2!)
from P{X=1}=P{X=2}, Solution :λ=2.
therefore P(X=4)=24×e-2÷4!=2/(3e2).
6 set up X~N(3,4), Try to ask for :(1)P{|X|>2};(2)P{X>3}.
Explain :(1)
(2) because N(3,4) About the mean 3 symmetry , therefore P{X>3}=1/2.
7 The life of electronic tubes produced by a factory X( In hours ) Obey expectations μ=160 Is a normal distribution , If required P{120<X<200}≥0.08, Allowable standard deviation σ What is the maximum ?
Explain : from
Available :Φ(40/σ)≥0.54, Look up the table :40/σ≥0.1004, so σ≤398.41. That is, the allowable standard deviation σ The maximum is 398.41.8 The number of errors in the proofreading of a book after typesetting X It follows a normal distribution N(200,400), seek :
(1) The number of errors does not exceed 230 Probability .
(2) The number of errors is 190~210 Probability between .
Explain :(1) The number of errors does not exceed 230 The probability of is :
(2) The number of errors is 190~210 The probability between is :
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