"assume x is a random variable with mean μ and standard deviation σ . find expressions for the mean and standard deviation of y
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Answer:
0.9910 = 99.10% probability that a sample of 170 steady smokers spend between $19 and $21
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, standard deviation of 5:
This means that
Sample of 170:
This means that
What is the probability that a sample of 170 steady smokers spend between $19 and $21?
This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.
X = 21
By the Central Limit Theorem
has a p-value of 0.9955
X = 19
has a p-value of 0.0045
0.9955 - 0.0045 = 0.9910
0.9910 = 99.10% probability that a sample of 170 steady smokers spend between $19 and $21