英文统计学题,An electronics firm claims that the proportion of defective units of a certain process is 5%.A buyer has a standard procedure of inspecting 15 units selected randomly from a large lot.On a particular occasion, the buyer found five

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英文统计学题,An electronics firm claims that the proportion of defective units of a certain process is 5%.A buyer has a standard procedure of inspecting 15 units selected randomly from a large lot.On a particular occasion, the buyer found five

英文统计学题,An electronics firm claims that the proportion of defective units of a certain process is 5%.A buyer has a standard procedure of inspecting 15 units selected randomly from a large lot.On a particular occasion, the buyer found five
英文统计学题,
An electronics firm claims that the proportion of defective units of a certain process is 5%.
A buyer has a standard procedure of inspecting 15 units selected randomly from a large lot.
On a particular occasion, the buyer found five defective items.

a) What is the probability of this occurrence, given that there is a 5% defect rate?
b) Use Minitab to find the probability that at least five defective items are found.

c) Given the answers to parts a) and b), do you think that the firm’s claim is reasonable?







Question 2

The probability that the noise level of a wide–band amplifier will exceed 1.5dB is 0.1. What
is the probability that in a set of 20 such amplifiers:

a)  Two of the amplifiers will exceed 1.5dB;

b)  At most two of the amplifiers will exceed 1.5dB;

c)  Between two and five of the amplifiers amplifiers will exceed 1.5dB (use the minitab
output below).

            Cumulative Distribution Function
            Binomial with n = 20 and p = 0.1

x P(X<=x)

1  0.391747

2  0.676927

4  0.956826

5  0.988747 

英文统计学题,An electronics firm claims that the proportion of defective units of a certain process is 5%.A buyer has a standard procedure of inspecting 15 units selected randomly from a large lot.On a particular occasion, the buyer found five
1.
a)查binomial的pmf表,找到对应的p=0.05,n=15,查出P(x=5)对应的概率.
b)minitab没发言时,其实就是1-P(x=0)-P(x=1)-P(x=2)-P(x=3)-P(x=4)
c)估计这个claim比较扯,看概率吧,如果a),b)的概率都是类似0.01这样的,可以说不靠谱,小概率事件.
2.
a)同上a),查表,n=20,p=0.1,查P(x=2)
b)P(x=0)+P(x=1)+P(x=2)
c)注意题目给的是CDF,所以0.988747-0.391747=0.597
GL