The set of life spans of an appliance is normally distributed with a mean [tex]$\mu=48$[/tex] months and a standard deviation [tex]$\sigma=$[/tex] 8 months. What is the life span of an appliance that has a [tex]$z$[/tex]-score of -3?
Answer (2)
Use the z-score formula: z = σ x − μ .
Plug in the given values: − 3 = 8 x − 48 .
Solve for x : x = 48 + ( − 3 ) ( 8 ) .
Calculate the value of x : x = 24 months.
Explanation
Understand the problem and provided data We are given that the life spans of an appliance are normally distributed with a mean μ = 48 months and a standard deviation σ = 8 months. We are asked to find the life span of an appliance that has a z -score of -3.
Recall the z-score formula The z -score formula is given by: z = σ x − μ where: x is the life span of the appliance, μ is the mean life span, σ is the standard deviation.
Plug in the given values We are given z = − 3 , μ = 48 , and σ = 8 . We need to solve for x . Plugging in the given values, we have: − 3 = 8 x − 48
Multiply both sides by 8 To solve for x , we multiply both sides of the equation by 8: − 3 × 8 = x − 48 − 24 = x − 48
Add 48 to both sides Now, we add 48 to both sides of the equation: − 24 + 48 = x x = 24
State the final answer Therefore, the life span of an appliance with a z -score of -3 is 24 months.
Examples
Understanding z-scores is crucial in many real-world applications. For example, in quality control, manufacturers use z-scores to determine if a product's specifications are within acceptable limits. If the z-score for a product's dimension is too high or too low, it indicates a deviation from the norm, and the manufacturing process may need adjustment. Similarly, in finance, z-scores can help assess the risk associated with an investment by measuring how far its returns deviate from the average.
The life span of an appliance with a z-score of -3 is 24 months, calculated using the z-score formula. This indicates that the appliance has a life span significantly below the average of 48 months. The calculation involves solving the equation formed by substituting the known values into the z-score formula.
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