What statistics formula do i need to solve this?

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?

What statistics formula do i need to solve this?
You need to find your z-scores.





a) z = (1970-1820)/120 = 150/120 = 1.25 %26lt;= find this in the standard normal table for get the area under the curve (the percentage)


It gives you .3944, so 39.44%


b) This will be the amount to the right of the part given in a.


So, .50 - .3944 = .1056 = 10.56%


c) z = (1600-1820)/120 = -220/120 = -1.8333


You are looking for the percentage to the left of this so your answer will be: .50 - the number given in the table for 1.83


d) This one you need to look in the values of the table to find where .25 occurs, which is about .675. This is the z-score. Use it to find the maximum number of passengers for cruises with the fewest 25% of passengers.


-.675 = (x-1820)/120
Reply:For any normal random variable X with mean μ and standard deviation σ, can be translated it to standard units so you can use the z tables by z = (X - μ) / σ








a) P(1820 %26lt; X %26lt; 1970) = P(0 %26lt; Z %26lt; 1.25)


= P(Z %26lt; 1.25) - P(Z %26lt; 0)


= 0.8944 - 0.5000


= 0.3943502





b) P(X %26gt; 1970) = 1-P(X%26lt;1970) = 1 - 0.8944 = 0.1056





c) P(X%26lt; 1600) = P(Z %26lt; -1.83) = 0.0336





d) P(Z %26lt; z) = 0.25


z = -0.6745


z = (X - 1820)/120


X = 1820 - 0.6745 * 120


X = 1739.06


X = 1740
Reply:I'd suggest you try Avril Lapoissons semi-distributed polynomial bifunctionalisation of the nth order polynomial.