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?
Try using the computional tool at this website:
http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html
Enter the mmean and the std. dev., and the limits of the integration interval, and then click "enter". Here are the answers I get:
(a) 39.4%
(b) 10.6%
(c) 3.3%
(d) 1739 passengers or less make up 25% of the sailings
Thank you
To answer these questions, we need to use the properties of the normal distribution. The mean number of passengers per cruise is given as 1,820, and the standard deviation is 120.
a. To find the percentage of cruises that will have between 1,820 and 1,970 passengers, we need to find the area under the normal distribution curve between those two values.
Step 1: Convert the values into z-scores. The formula for z-score is z = (x - mean) / standard deviation.
For 1,820 passengers: z = (1,820 - 1,820) / 120 = 0
For 1,970 passengers: z = (1,970 - 1,820) / 120 ≈ 1.25
Step 2: Use a Z-table or a statistical software to find the corresponding areas under the standard normal distribution curve for the z-scores calculated in Step 1.
The area to the left of z = 0 is 0.5000.
The area to the left of z = 1.25 is approximately 0.8944.
Step 3: Find the difference between these two areas to get the percentage of cruises that will have between 1,820 and 1,970 passengers.
Percentage = (0.8944 - 0.5000) * 100 ≈ 39.44%
Therefore, approximately 39.44% of the cruises will have between 1,820 and 1,970 passengers.
b. To find the percentage of cruises that will have 1,970 passengers or more, we need to find the area to the right of the z-score corresponding to 1,970 passengers.
Using the z-score calculated previously: z = 1.25
Using a Z-table or statistical software, we can find the area to the left of z = 1.25 as approximately 0.8944.
The area to the right of z = 1.25 is given by: 1 - 0.8944 = 0.1056
Therefore, approximately 10.56% of the cruises will have 1,970 passengers or more.
c. To find the percentage of cruises that will have 1,600 or fewer passengers, we need to find the area to the left of the z-score corresponding to 1,600 passengers.
Step 1: Convert 1,600 passengers into a z-score.
z = (1,600 - 1,820) / 120 ≈ -1.83
Step 2: Using a Z-table or statistical software, find the area to the left of z = -1.83.
The area to the left of z = -1.83 is approximately 0.0336.
Therefore, approximately 3.36% of the cruises will have 1,600 or fewer passengers.
d. To find the number of passengers on the cruises with the fewest 25% of passengers, we calculate the z-score corresponding to the 25th percentile.
Step 1: Determine the z-score that corresponds to the 25th percentile.
Using a Z-table or statistical software, the z-score corresponding to the 25th percentile is approximately -0.674.
Step 2: Convert the z-score back into the actual number of passengers.
Number of passengers = (z-score * standard deviation) + mean
Number of passengers = (-0.674 * 120) + 1,820 ≈ 1,748.48
Therefore, the cruises with the fewest 25 percent of passengers will have approximately 1,748 passengers.