On 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 and 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?

Using the normal distribution, it is found that there is a 0.0082 = 0.82% probability of such an accomplishment.

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

In this problem:

The probability is the p-value of Z when X = 33, hence:

has a p-value of 0.0082.

0.0082 = 0.82% probability of such an accomplishment.

A similar problem is given at link

0.0082 or 0.82%

Step-by-step explanation:

Given:

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:

Plug in the values and solve for 'z'. This gives,

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%.

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

(23*2)-12=x     x=34

step-by-step explanation: