Question

Cruise Burmuda marketers want to make a “soft promise”to customers that they won’t spend more than $200 at tour destinations. Apply the NORM.DIST function in Excel to find the probability that a Cruise Burmuda tourist spends less than $200 for “cruise tourist spending”:
- What is the probability that Cruise Burmuda customers will spend less than $200 for “cruise tourist spending”?
- What is the probability that Cruise Burmuda customers will spend more than $200 for “cruise tourist spending”?
- Is this good enough to make a promise to customers?

Assume the mean is $185 and the standard deviation is $15.

Answer 1

given that

mean = 185 and standard deviation= 15

(A) using NORM.DIST

setting x = 200, mean = 185, sd = 15 and true

we get

=NORM.DIST(200,185,15,true)

= 0.8413

(B)

using NORM.DIST

setting x = 200, mean = 185, sd = 15 and true

we get

P(x>200) = 1 - NORM.DIST(200,185,15,true)

= 1 - 0.8413

= 0.1587

(C) No, this is not good enough because the probability of spending more than $200 is not unusual or less than 5%. So, it is usual to spend more than $200.

Therefore, it is not good enough