Statistics 8th Edition Jia Junping Chapter 5 probability and probability distribution

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 ：