Evaluate (–1)8 + (–1)7 + –16 + –14 – (–1)2.
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Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9
We have to find the probability that in a randomly selected week the number of burglaries is at least three.
P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........
= 1 - P(X < 3)
= 1 - [ P(X=2) + P(X=1) + P(X=0)]
The Poisson probability at X=k is given by
P(X=k) =
Using this formula probability of X=2,1,0 with mean = 1.9 is
P(X=2) =
P(X=2) =
P(X=2) = 0.2698
P(X=1) =
P(X=1) =
P(X=1) = 0.2841
P(X=0) =
P(X=0) =
P(X=0) = 0.1495
The probability that at least three will become
P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]
= 1 - [0.2698 + 0.2841 + 0.1495]
= 1 - 0.7034
P(X ≥ 3 ) = 0.2966
The probability that in a randomly selected week the number of burglaries is at least three is 0.2966
Answer:
Step-by-step explanation:
Hello!
The objective of this experiment is to compare two compounds, designed to reduce braking distance, used in tire manufacturing to prove if the braking distance of SUV's equipped with tires made with compound 1 is shorter than the braking distance of SUV's equipped with tires made with compound 2.
So you have 2 independent populations, SUV's equipped with tires made using compound 1 and SUV's equipped with tires made using compound 2.
Two samples of 81 braking tests are made and the braking distance was measured each time, the study variables are determined as:
X₁: Braking distance of an SUV equipped with tires made with compound one.
Its sample mean is X[bar]₁= 69 feet
And the Standard deviation S₁= 10.4 feet
X₂: Braking distance of an SUV equipped with tires made with compound two.
Its sample mean is X[bar]₂= 71 feet
And the Standard deviation S₂= 7.6 feet
We don't have any information on the distribution of the study variables, nor the sample data to test it, but since both sample sizes are large enough n₁ and n₂ ≥ 30 we can apply the central limit theorem and approximate the distribution of both variables sample means to normal.
The researcher's hypothesis, as mentioned before, is that the braking distance using compound one is less than the distance obtained using compound 2, symbolically: μ₁ < μ₂
The statistical hypotheses are:
H₀: μ₁ ≥ μ₂
H₁: μ₁ < μ₂
α: 0.05
The statistic to use to compare these two populations is a pooled Z test
Z ≈ N(0;1)
The rejection region if this hypothesis test is one-tailed to the right, so you'll reject the null hypothesis to small values of the statistic. The critical value for this test is:
Decision rule:
If > -1.648 , then you do not reject the null hypothesis.
If ≤ -1.648 , then you reject the null hypothesis.
Since the statistic value is greater than the critical value, the decision is to not reject the null hypothesis.
At a 5% significance level, you can conclude that the average braking distance of SUV's equipped with tires manufactured used compound 1 is greater than the average braking distance of SUV's equipped with tires manufactured used compound 2.
I hope you have a SUPER day!