Mallie
date
8. travel times journal found that the average per person cost of a
ound that the average per person cost of a 10-day trip along the pacific
coast, per person, is $1,015. this includes transportation, food, lodging, and entertainment.
a. if the data is normally distributed with standard deviation $198, find
ard deviation $198, find the percent of
vacationers who spent less than $1.200 per day. round to the nearest hundredth of a
percent.

82.38%

Step-by-step explanation:

Average cost = u = $ 1,015

Standard Deviation = = $ 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:

Using the values in this formula, we get:

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%

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.