The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $2000 and the standard deviation is $65.
What is the approximate percentage of buyers who paid less than $1870?
To find the approximate percentage of buyers who paid less than $1870, we need to calculate the z-score for this value and look up the corresponding area under the normal curve.
The z-score formula is:
z = (x - μ) / σ
where:
x = $1870 (the value we want to find the percentage for)
μ = $2000 (mean price paid)
σ = $65 (standard deviation)
Substituting the values into the formula:
z = (1870 - 2000) / 65
z = -0.2
Next, we need to use a standard normal distribution table or a calculator to find the area to the left of z = -0.2. The closest value we can find from the table is z = -0.22, which has an area of 0.4131.
However, this value corresponds to the area to the left of z, and we are interested in the area to the right (since we want the percentage of buyers who paid less than $1870). Thus, we subtract the area from 1:
P(Z < -0.2) ≈ 1 - 0.4131
P(Z < -0.2) ≈ 0.5869
To convert this to a percentage, we multiply by 100:
Percentage of buyers who paid less than $1870 ≈ 0.5869 * 100 ≈ 58.69%
Therefore, approximately 58.69% of buyers paid less than $1870 based on the normal distribution.