Respuesta :
Answer:
1.32% probability that a randomly selected 100 customers will spend an average of less than $50 on dinner
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 54, \sigma = 18, n = 100, s = \frac{18}{\sqrt{100}} = 1.8[/tex]
Based on this information, what is the probability that a randomly selected 100 customers will spend an average of less than $50 on dinner?
This probability is the pvalue of Z when X = 50. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{50 - 54}{1.8}[/tex]
[tex]Z = -2.22[/tex]
[tex]Z = -2.22[/tex] has a pvalue of 0.0132.
1.32% probability that a randomly selected 100 customers will spend an average of less than $50 on dinner
