Question

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?

Answer 1

Answer:

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:

$z=\frac{x-\mu}{\sigma}$

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

$z=\frac{33-45}{5}=-2.4$

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

Now, we need the probability $P(x\leq 33)=P(z\leq -2.4)$.

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%.

Answer 2

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Mean of 45 seconds, hence $\mu = 45$.
• Standard deviation of 5 seconds, hence $\sigma = 5$.

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{33 - 45}{5}$

$Z = -2.4$

$Z = -2.4$ has a p-value of 0.0082.

0.0082 = 0.82% probability of such an accomplishment.

A similar problem is given at https://brainly.com/question/24663213