Solve for b: 16b = 12
1 answer:
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Answer:
The p-value of the test is 0.9918 > 0.05, which means that we fail to reject , as we do not have enough evidence to say that defect rate of machine is smaller than 3%.
Step-by-step explanation:
The null and alternate hypothesis are:
H0:p≥0.03
Ha:p<0.03
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
0.03 is tested at the null hypothesis:
This means that
28 out of 600 items produced by the machine were found defective.
This means that
Value of the test-statistic:
P-value of the test:
The p-value of the test is the probability of finding a sample proportion below 0.0467, which is the p-value of z = 2.4.
Looking at the z-table, z = 2.4 has a p-value of 0.9918.
The p-value of the test is 0.9918 > 0.05, which means that we fail to reject , as we do not have enough evidence to say that defect rate of machine is smaller than 3%.
Answer:
(a) The histogram is shown below.
(b) E (X) = 4.2
(c) SD (X) = 2.73
Step-by-step explanation:
Let <em>X</em> = <em>r</em><em> </em>= a driver will tailgate the car in front of him before passing.
The probability that a driver will tailgate the car in front of him before passing is, P (X) = <em>p</em> = 0.35.
The sample selected is of size <em>n</em> = 12.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 12 and <em>p</em> = 0.35.
The probability function of a binomial random variable is:
(a)
For <em>X</em> = 0 the probability is:
For <em>X</em> = 1 the probability is:
For <em>X</em> = 2 the probability is:
Similarly the remaining probabilities will be computed.
The probability distribution table is shown below.
The histogram is also shown below.
(b)
The expected value of a Binomial distribution is:
The expected number of vehicles out of 12 that will tailgate is:
Thus, the expected number of vehicles out of 12 that will tailgate is 4.2.
(c)
The standard deviation of a Binomial distribution is:
The standard deviation of vehicles out of 12 that will tailgate is:
Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.
