Joan is buying airline tickets for her family. Each airline ticket costs $232 with a baggage fee of $13. The airline charges a b
2 answers:
You might be interested in
Answer:
0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 20 hours, standard deviation of 6:
This means that
Sample of 150:
This means that
What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours?
This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.5. So
X = 21
By the Central Limit Theorem
has a p-value of 0.9793
X = 19.5
has a p-value of 0.1539
0.9793 - 0.1539 = 0.8254
0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours