The probability of a plane crashing on a single trip in 1989 was 0.00000165. Find the approximate probability that in 50,000,000 flights there will be fewer than 90 crashes. Round your answer to four decimal places. Round Z to two decimal places.
A)0.5332
B) 0.7804
C) 0.7784
D) 0.5342
E) 0.7794

Question

The probability of a plane crashing on a single trip in 1989 was 0.00000165. Find the approximate probability that in 50,000,000 flights there will be fewer than 90 crashes. Round your answer to four decimal places. Round Z to two decimal places. ​
A)0.5332
B) 0.7804
C) 0.7784
D) 0.5342
E) 0.7794

Quizplus · Accepted Answer

This problem can be approached using the normal approximation to the binomial distribution. The mean (μ) of crashes in 50,000,000 flights can be calculated as n*p = 50,000,000 * 0.00000165 = 82.5. The standard deviation (σ) is sqrt(n*p*(1-p)) = sqrt(50,000,000*0.00000165*(1-0.00000165)) ≈ 9.07. To find the probability of fewer than 90 crashes, we calculate the Z-score for 89.5 (using continuity correction) as (89.5 - 82.5) / 9.07 ≈ 0.77. Looking up this Z-score in a standard normal distribution table or using a calculator gives a probability of approximately 0.7794.

The Probability of a Plane Crashing on a Single Trip