For a standard normal distribution, find the approximate value of P (z greater-than-or-equal-to negative 1. 25). Use the portion
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Answer:
71.123 mph ≤ μ ≤ 77.277 mph
Step-by-step explanation:
Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:
≤ μ ≤
Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.
the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.
So, if we replace m by 74.2, s by 5.3083, n by 10 and by 1.8331, we get that the 90% confidence interval for the mean speed is:
≤ μ ≤
74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077
71.123 ≤ μ ≤ 77.277