A set of 50 data values has a mean of 30 and a variance of 4.
I. Find the standard score (z) for a data value = 26.
II. Find the probability of a data value < 26.
I think the first part is z=(26-30)/4 = -4/4 = -1. I am not sure how to go about the second problem. I could use any help. Thanks
1. Z =( score-mean)/SD
Variance = SD squared
2. Same equation, but find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to Z.
To find the probability of a data value less than 26, we need to use the standard normal distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1.
To solve the second problem, follow these steps:
Step 1: Standardize the data value.
To find the standard score (z) for a data value of 26, you can use the formula: z = (x - μ) / σ, where x is the data value, μ is the mean, and σ is the standard deviation.
z = (26 - 30) / 4
z = -4 / 4
z = -1
So, the standard score for the data value 26 is -1.
Step 2: Use a Z-table or calculator.
The Z-table provides the area/probability under the standard normal distribution curve for a given z-score. However, tables often provide the area/probability to the left of the z-score, so we need to take that into account.
To find the probability of a data value less than 26, we need to find the area under the curve to the left of -1 on the standard normal distribution.
Using a Z-table or a statistical calculator, you can find that the area/probability to the left of -1 is approximately 0.1587. This means that there is a 15.87% probability of randomly selecting a data value less than 26 from the given set.
In summary, the probability of a data value less than 26 is approximately 0.1587 or 15.87%.