The average homicide rate for the cities and towns in a state is 10 per 100,000 population with a standard deviation of 2. If the variable is normally distributed, what is the probability that a randomly selected town will have a homicide rate greater than 14?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
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The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 8 minutes. Find the probability that a randomly selected passenger has a waiting time greater thangreater than 2.25 minutes.
Find the probability that a randomly selected passenger has a waiting time greater thangreater than 2.25 minutes.
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find the probability that a randomly selected passenger has a waiting time greater than 2.25
To answer this question, we can use the concept of z-scores and the standard normal distribution.
First, let's calculate the z-score for a homicide rate of 14:
z = (x - μ) / σ
Where:
- x is the observed value (14 in this case)
- μ is the mean (10 in this case)
- σ is the standard deviation (2 in this case)
Plug in the values:
z = (14 - 10) / 2 = 4 / 2 = 2
Next, we need to find the probability corresponding to this z-score. We can refer to a z-table or use a calculator to find the probability.
Since we want the probability of a homicide rate greater than 14, we need to find the probability associated with the right tail of the distribution. This can be calculated by subtracting the cumulative probability of the left tail from 1.
Using a z-table or calculator, we find that the cumulative probability associated with a z-score of 2 is approximately 0.9772.
Now, subtracting this value from 1 gives us the probability of a homicide rate greater than 14:
P(x > 14) = 1 - 0.9772 = 0.0228
Therefore, the probability that a randomly selected town will have a homicide rate greater than 14 is approximately 0.0228 or 2.28%.