Help I’m doing a test.... what is 3(2x-9) =
2 answers:
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Answer:
64.65% probability of at least one injury commuting to work in the next 20 years
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Each day:
Bikes to work with probability 0.4.
If he bikes to work, 0.1 injuries per year.
Walks to work with probability 0.6.
If he walks to work, 0.02 injuries per year.
20 years.
So
Either he suffers no injuries, or he suffer at least one injury. The sum of the probabilities of these events is decimal 1. So
We want . Then
In which
64.65% probability of at least one injury commuting to work in the next 20 years
Answer:
a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.
b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.
Step-by-step explanation:
Question a:
We have to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Z-table as such z has a p-value of .
That is z with a p-value of , so Z = 1.96.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.
The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.
The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.
Question b:
The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.