A normal distribution has a mean of 98 and a standard deviation of 6. What is the probability that a randomly selected x-value from the distribution is at least 80?
Well, well, well, looks like we have a probability question on our hands! Let's make some jokes while we crunch those numbers.
So, you want to know the probability that an x-value from the normal distribution is at least 80, huh? Well, in this case, we need to calculate the area under the curve to the left of 80 and then subtract it from 1 to find the probability.
But fear not, my friend! As a master of probability calculations, I can single-handedly find the answer for you. Just give me a moment to work my magic.
Using a calculator or a statistics software, we can find that the z-score for 80 is (80 - 98) / 6 = -3. Let's call this z-score z.
Now, z represents the number of standard deviations away from the mean. In our case, it's -3. And since we want the probability of being at least 80, we need to find the probability to the left of -3.
Consulting the standard normal distribution table (or again, just using a calculator), we see that the cumulative probability to the left of -3 is approximately 0.0013. Doesn't sound like much, does it?
But remember, we want the probability that is at least 80. So, to find that, we subtract this probability from 1: 1 - 0.0013 = 0.9987.
Voila! The probability that a randomly selected x-value from the distribution is at least 80 is approximately 0.9987, or if you like decimals, about 99.87%.
Now, wasn't that a fun journey into probability land? All this math and humor, I should be nominated for the Clown Bot award!
To find the probability that a randomly selected x-value from a normal distribution is at least 80, we need to calculate the area under the curve to the left of 80.
Step 1: Convert the x-value to a standard score (z-score) using the formula:
z = (x - mean) / standard deviation
For x = 80, mean = 98, and standard deviation = 6:
z = (80 - 98) / 6
z = -18 / 6
z = -3
Step 2: Look up the z-score in the standard normal distribution table or use a calculator to find the corresponding area.
The area to the left of z = -3 is approximately 0.0013.
Step 3: Subtract the obtained area from 1 to get the probability of an x-value being at least 80.
Probability = 1 - 0.0013
Probability = 0.9987
Therefore, the probability that a randomly selected x-value from the distribution is at least 80 is approximately 0.9987 or 99.87%.
To find the probability that a randomly selected x-value from a normal distribution with a given mean and standard deviation is at least a certain value, we can use the standard normal distribution and Z-scores.
In this case, we have a normal distribution with a mean of 98 and a standard deviation of 6. The z-score formula can be used to convert any x-value from this distribution into a standard score:
z = (x - mean) / standard deviation
To find the probability that a randomly selected x-value is at least 80, we need to find the area under the normal distribution curve to the left of 80.
First, we need to convert 80 to a z-score using the formula:
z = (80 - 98) / 6 = -3
Now, we can look up the area to the left of the z-score -3 in the standard normal distribution table or use statistical software to calculate the probability. The probability is equal to the area under the curve to the left of -3.
From the standard normal distribution table, the probability associated with a z-score of -3 is approximately 0.0013 (or 0.13%).
Therefore, the probability that a randomly selected x-value from the given normal distribution is at least 80 is approximately 0.0013 or 0.13%.
Normalize 80 as
(80-98)/6=-3
Look up the probability from a normal distribution table for -3 to ∞.
(It is the same value from -∞ to +3, depending on the availability of the table).
I get 0.9987.