Super stumped... Help please.
The average distance geese fly south in the winter is approximately 380 miles with a standard deviation of 60 miles. Assuming that the distances are normally distributed, find the probability that a randomly selected goose migrates a distance of:
a. Less than 300 miles
b. Between 320 a 440 miles
c. Greater than 460 miles
300 miles is -80/60 = -1.33σ from the mean
320-400 is -1.00σ to + +0.33σ
>460 miles is >1.33σ
So, just go tour standard Z table and read off the values. For the first, you want the area of the left tail at least 1.33σ from the mean: 0.091
A handy Z-table grapher and solver can be found at
http://davidmlane.com/hyperstat/z_table.html
Very nice. that Z-grapher helped. :)PZ
To find the probabilities in this problem, we will use the standard normal distribution table or a calculator such as a Z-table or a graphing calculator. However, since it is not mentioned, we will assume that the distances follow a normal distribution.
First, we need to standardize the values using the z-score formula:
z = (x - μ) / σ
Where:
- x is the value we want to find the probability for,
- μ is the mean (average distance), and
- σ is the standard deviation.
a. To find the probability that a randomly selected goose migrates a distance of less than 300 miles, we need to calculate the z-score for 300 miles:
z = (300 - 380) / 60
z = -1.33
Next, we need to find the probability corresponding to this z-score. Looking up the z-score in the standard normal distribution table or using a calculator will give us the probability.
b. To find the probability that a randomly selected goose migrates a distance between 320 and 440 miles, we need to calculate the z-scores for both distances.
For 320 miles:
z = (320 - 380) / 60
z = -1
For 440 miles:
z = (440 - 380) / 60
z = 1
We now have two z-scores, -1 and 1. To find the probability between these two distances, we need to find the area between these two z-scores using the standard normal distribution table or a calculator.
c. To find the probability that a randomly selected goose migrates a distance greater than 460 miles, we need to calculate the z-score for 460 miles:
z = (460 - 380) / 60
z = 1.33
Again, we can use the z-score to find the corresponding probability using the standard normal distribution table or a calculator.
Please note that the actual values obtained from the standard normal distribution table or calculator may differ slightly depending on the specific tool or method used.