The life span of a DVD player produced by one major company is known to be normally distributed with a mean of 6.2 years. If 4.01% of these DVD players have a life span of more than 8 years, what is the standard deviation of the DVD player’s life span?
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 (.0401) and its Z score. Insert values in above equation to get SD.
To find the standard deviation of the DVD player's life span, we need to use the information provided and the properties of a normal distribution.
Let's break down the problem:
1. We are given that the life span of DVD players produced by the major company is normally distributed with a mean of 6.2 years.
2. We are also given that 4.01% of these DVD players have a life span of more than 8 years.
To find the standard deviation, we need to use the Z-score formula.
The Z-score formula is given by:
Z = (x - μ) / σ
Where:
Z is the Z-score,
x is the observed value,
μ is the mean, and
σ is the standard deviation.
Since we know the mean (μ) is 6.2 years, and we want to find the standard deviation (σ), we need to find the value of Z.
To find Z, we can use the Z-table or a statistical calculator. The Z-score corresponding to 4.01% (0.0401) is approximately 1.764.
So, using the Z-score formula, we can write the equation as:
1.764 = (8 - 6.2) / σ
Now, we can solve for the standard deviation (σ):
1.764σ = 8 - 6.2
1.764σ = 1.8
Divide both sides of the equation by 1.764:
σ ≈ 1.8 / 1.764
σ ≈ 1.019
Therefore, the standard deviation of the DVD player's life span is approximately 1.019 years.