A flagpole in front of the Post Office was 38 feet tall. It was hit by a truck and now it's only 42 inches tall. What length of
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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
Answer:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.
Step-by-step explanation:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.