The scores on a standardized test are normally distributed with the mean of 750 and standard deviation 70. Find the probability that a score is more than 890.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
The scores on a standardized test are normally distributed with the mean of 750 and standard deviation 70. Find the probability that a score is more than 890.
This is from my Math Models class. I am not taking Statistics. Please someone show me how to do answer this-I don't just want an answer. I want the steps!!
To find the probability that a score is more than 890, we need to convert this score into a standardized z-score and then find the corresponding area under the normal distribution curve.
The z-score formula is given by:
z = (x - μ) / σ
Where:
x = score
μ = mean
σ = standard deviation
In this case, the score (x) is 890, the mean (μ) is 750, and the standard deviation (σ) is 70.
Using the formula, we calculate the z-score:
z = (890 - 750) / 70
z = 2
Now, we need to find the area to the right of z = 2 under the normal distribution curve.
Using a standard normal distribution table or a statistical calculator, we find that the area to the right of z = 2 is 0.0228.
Therefore, the probability that a score is more than 890 is 0.0228, or 2.28%.
To find the probability that a score is more than 890, we can use the standard normal distribution.
Step 1: Standardize the score
To standardize a value, we subtract the mean and divide by the standard deviation. In this case, to standardize a score of 890, we use the formula:
Z = (X - μ) / σ
where X is the score, μ is the mean, and σ is the standard deviation.
Z = (890 - 750) / 70
Z = 140 / 70
Z = 2
Step 2: Find the probability
Once we have the standardized score (Z-value), we can use a standard normal distribution table or a calculator to find the probability.
The probability of a score being more than 890 can be represented as P(Z > 2).
Using a standard normal distribution table, we can find the corresponding probability. Looking up a Z-value of 2 in the table, we find that the probability is approximately 0.0228.
Therefore, the probability that a score is more than 890 is approximately 0.0228, or 2.28%.