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!!
this just involves using the Z table.
890 is 140 away from the mean.
That's 140/70 = 2σ
So, find where Z>2. That's your probability. To play around some with this, visit
http://davidmlane.com/hyperstat/z_table.html
To find the probability that a score on a standardized test is more than 890, you can use the normal distribution formula and the z-score.
The first step is to calculate the z-score for the value 890. The z-score formula is:
z = (x - μ) / σ
Where:
x is the value you are interested in (890 in this case)
μ (mu) is the mean of the distribution (750 in this case)
σ (sigma) is the standard deviation of the distribution (70 in this case)
Plugging in the values, we can calculate the z-score:
z = (890 - 750) / 70
z = 5.2
The next step is to find the probability associated with the z-score. You can use a standard normal distribution table or a calculator to find this value.
For example, using a standard normal distribution table, the probability corresponding to a z-score of 5.2 is approximately 0.99999987.
Since we want the probability that a score is more than 890, we need to find the complement of this probability (1 - probability). So the final answer would be:
1 - 0.99999987 ≈ 0.00000013
Therefore, the probability that a score is more than 890 is approximately 0.00000013 or close to 0%.