Probability

Slick

New Member
Question-having a hard time understanding this-help, I just do not get it...
The number of passengers on the Carnival Sensation during one-week cruises in the
Caribbean follows the normal distribution. The mean number of passengers per cruise is
1,820 and the standard deviation is 120.
a. What percent of the cruises will have between 1,820 and 1,970 passengers?
b. What percent of the cruises will have 1,970 passengers or more?
c. What percent of the cruises will have 1,600 or fewer passengers?
d. How many passengers are on the cruises with the fewest 25 percent of passengers? Mean Joe

TS Contributor
You'll need a normal table.
Let X = number of passengers. Given: X ~ N(mean=1820, var=120^2).

In general, for some number N
P[X < N] = P[Z < (N-1820)/120], then use the normal table (it gives a decimal number, you convert to a %)
Similarly,
P[N < X] = P[(N-1820)/120 < Z] = 1 - P[Z < (N-1820)/120], which you use a normal table to find.

Use this for part a:
P[M < X < N] = P[X < N] - P[X < M]

Keo

New Member
Hint:
Remember that on your tables you can find different areas under the curve (and hence percentages) for any given z score, and you can convert any number to z scores, since 1 SD = 1 z. i.e. 180 is equal to 1.5 z's.
the distance between 1,820 and 1,970 is 150. What is that in z scores?
Good luck.

Dragan

Super Moderator
Question-having a hard time understanding this-help, I just do not get it...
It's an exercise on how to make use of the standard normal distribution.

For example, part (a):

z=(x-Mu)/Sigma = (1970-1820)/120 = 1.25.

Thus, using the textbook, we find that proportion of passengers between 0 and 1.25 standard deviations is p=.3944.

Hence, 39.44% of the cruises will contain passengers that number between 1970 and 1820.

Mkay.