Answer:
82.38%
Step-by-step explanation:
Average cost = u = $ 1,015
Standard Deviation = [tex]\sigma[/tex] = $ 198
Its given that data is normally distributed. We have to find what percentage of vacationers spend less than $ 1200.
Since the data is normally distributed, we can use the concept of z score to answer this question.
We have to find:
Probability ( Spending < 1200)
In symbolic form, this can be represented as:
P(X < 1200)
The formula for the z score is:
[tex]z=\frac{x-u}{\sigma}[/tex]
Using the values in this formula, we get:
[tex]z=\frac{1200-1015}{198}=0.93[/tex]
Thus,
P(X < 1200) is equivalent to P(z < 0.93)
From the z table we can find the probability of z scores being less than 0.93 to be 0.8238
Thus,
P(z < 0.93) = 0.8238
Since,
P(X < 1200) = P(z < 0.93), we can conclude:
The percent of Vacationers who spent less than $1,200 is 0.8238 or 82.38%
