Question

Mallie

Answer 1

Answer:

82.38%

Step-by-step explanation:

Average cost = u = $ 1,015

Standard Deviation = $\sigma$ = $ 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:

$z=\frac{x-u}{\sigma}$

Using the values in this formula, we get:

$z=\frac{1200-1015}{198}=0.93$

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%