Answer:
57.62% probability that the proportion of people who will order coffee with their meal is within 2% of the population mean.
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 normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, 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.
For a proportion p in a sample of size n, we have that
In this question:
So
What is the probability that the proportion of people who will order coffee with their meal is within 2% of the population mean?
This is the pvalue of Z when X = 0.9 + 0.02 = 0.92 subtracted by the pvalue of Z when X = 0.9 - 0.02 = 0.88. So
X = 0.92
By the Central Limit Theorem
has a pvalue of 0.7881
X = 0.88
has a pvalue of 0.2119
0.7881 - 0.2119 = 0.5762
57.62% probability that the proportion of people who will order coffee with their meal is within 2% of the population mean.
