Question

When the original Star Wars movie came out (1977), there was much excitement about the movie. I am a huge Star Wars fan. Please solve the following questions - you must show ALL of your work to receive credit for the question.

1. you must write the problem in probability notation.

2. you must draw the distribution and mark off everything that is given in the problem.

3. you must use the correct formula to solve the problem and show your work.

4. you must write out the final answer for the problem.

a. On the average, it takes Han Solo 45 seconds to check the coordinates and make the jump into hyperspace. The standard deviation on this important task is 5 seconds. When Han Solo, Chewbacca and their passengers are leaving for Alderaan they make the jump in 33 seconds or less. What is the probability of such an accomplishment?

b. In a space bar, there were 14 storm troopers, 3 Wookies, 9 humans, and 2 scriptwriters. An Android entered, fired a shot, and hit someone in the cheek. What is the probability that a scriptwriter was hit?

c. Jawas, those jewel-eyes, hooded collectors of robots and scrap, live in the desert and travel by sandcrawler. Their height is normally distributed with a mean of four feet and a standard deviation of 3 inches. The escape exit on the sandcrawler is 46 inches high. What proportion of the Jawas must duck when they use the escape exit?

Answer 1

a. Mean of jump time (μ) = 45 s

Standard deviation (σ) = 5 s

Time for jump required for accomplishment (x) = 33 s

The distribution is Normal Distribution.

So, first, we will find the z-score of the distribution using the formula:

z = (x-μ)/σ

z = 33-45/5 = -2.4

So, the z-score of the distribution is -2.4.

Now, we need the probability. From the normal distribution table for z-score equal to -2.4, the value of the probability is 0.0082 or 0.82%. You may use z score table to find the probability.

Therefore, the probability of making a jump in 33 seconds or less is 0.0082 or 0.82%.

b. Number of scriptwriters = 2

total = 14 + 3 + 9 + 2 =28

The probability that a scriptwriter was hit = 2/28 = .0714

7.14% probability that it was a scriptwriter hit.

c. Mean height(μ) = 4 feet = 48 inches

Standard deviation (σ) = 3 inches

x= 46 inches

Calculate the Z score: z = 46 – 48/3 = -0.67

Again, looking at the z table, This corresponds with 0.74857.

So 74.86% must duck when they use the escape exit.