1. A consumer survey indicates that the average household spends μ = $155 on groceries each week. The distribution of spending amounts is normal with a standard deviation of σ = $25. Based on this distribution: What is the probability of randomly selecting a family that spends less than $110 per week? (answers can be provided in decimal or percentage format) 2. A consumer survey indicates that the average household spends μ = $155 on groceries each week. The distribution of spending amounts is normal with a standard deviation of σ = $25. Based on this distribution: What is the z-score that corresponds to a family that spends less than $110 per week? z < ________?
Added by Albert B.
Step 1
Step 1: Calculate the z-score for a family that spends less than $110 per week using the formula: \[ z = \frac{X - \mu}{\sigma} \] where: \( X = $110 \) (weekly expenditure), \( \mu = $155 \) (mean expenditure), \( \sigma = $25 \) (standard deviation). Show more…
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