A normal distribution has a mean of 100 and standard deviation of 20. What is the
probability of randomly selecting a score less than 130 from this distribution?
a, p=0.9032
b, p=0.9332
c, p=0.0968
d, p=0.0668
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
First notice they gave you the mean, standard deviation, and an X value. Meaning you can you the formula
I would also draw a distribution picture where the number will end up being.
X = Z * (standard deviation) + (mean)
Plug in the numbers:
130 = Z (20) + 100
Subtract 100 FROM 130:
30 = Z (20)
Divide:
30/20
Your Z-score is:
Z = 1.5
*** Since you are looking for a probability LESS than 130, the probability will be on the LEFT side of the distribution. This means you have to look on your "Unit Normal Table" for a Z score of 1.5 and in the "B" (proportion body) column.
The answer is 0.9332
**Note every table may vary, so look to see what you B column says for the Z score of 1.5.
To find the probability of randomly selecting a score less than 130 from a normal distribution with a mean of 100 and a standard deviation of 20, you can use the standard normal distribution table or a calculator.
First, calculate the z-score, which represents how many standard deviations a particular score is from the mean:
z = (130 - 100) / 20 = 30 / 20 = 1.5
Using the standard normal distribution table or a calculator, find the probability corresponding to the z-score of 1.5. Looking up the z-score of 1.5 in the table, you will find that the probability is approximately 0.9332.
Therefore, the correct answer is B, p=0.9332.
To find the probability of randomly selecting a score less than 130 from a normal distribution with a mean of 100 and standard deviation of 20, we need to use a standard normal distribution table or a calculator.
The first step is to standardize the score by calculating the z-score. The z-score formula is:
z = (x - μ) / σ
Where:
x is the value we are interested in (130 in this case),
μ is the mean of the distribution (100 in this case), and
σ is the standard deviation of the distribution (20 in this case).
So, let's calculate the z-score:
z = (130 - 100) / 20
z = 30 / 20
z = 1.5
Once we have the z-score, we can find the corresponding probability from the standard normal distribution table or by using a calculator.
From the standard normal distribution table, we look for the value 1.5. The closest value we find is 1.0, which corresponds to a probability of 0.8413. We also look for the value 0.50, which corresponds to a probability of 0.5000. By subtract the two probabilities, we get the area under the curve from -∞ to 1.5.
The probability of randomly selecting a score less than 130 from this distribution is:
p = 0.8413 - 0.5000
p = 0.3413
Therefore, the correct answer is not in the given options.