Answer:
0.8808
Step-by-step explanation:
The objective is to calculate the probability of:
P(-1.20 ≤ z ≤ 2.64)
i.e. the probability that is the area between two z-values.
Thus;
P(-1.20 ≤ z ≤ 2.64) = P(z ≤ 2.64) - P( z ≤ 1.20)
From z table;
P(-1.20 ≤ z ≤ 2.64) = 0.9959 - 0.1151
P(-1.20 ≤ z ≤ 2.64) = 0.8808
We can see the shaded area under the standard normal curve from the image attached below.
The question is incomplete. Here is the complete qeustion.
In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained: 0.10 0.13 0.16 0.15 0.14 0.008 0.15
(a) Construct a 99% confidence interval for the mean nitrogen-oxide emissions of all cars.
(b) If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, based on the 99% confidence interval in (a), can we safely conclude that this requirement is being met?
Answer: (a) 0.089 ≤ μ ≤ 0.171
(b) No
Step-by-step explanation:
(a) To determine the confidence interval, first calculate the mean (X) and standard deviation (s) of the sample
X =
X = 0.13
s =
s = 0.029
The degrees of freedom is
N - 1 = 7 - 1 = 6
And since the confidence is of 99%:
α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005
The t-test statistics for is 3.707
(Value found in the t-distribution table)
Now, calculate Error:
E = .
E = 3.707.
E = 0.041
The interval will be:
0.13 - 0.041 ≤ μ ≤ 0.13+0.041
0.089 ≤ μ ≤ 0.171
(b) No, because according to the interval, the nitrode-oxide emissions range from 0.089 to 0.171, which is greater than required by EPA.