A consumer survey indicates that the average household spends μ = 170 on groceries each week. The distribution of spending amounts is approximately normal with a standard deviation of σ = 25. What is the probability of selecting a household that spends between $140 to $210 on groceries each week? What is the z-score that separates the top 20% of the distribution? What are the z-scores that separate the distribution in the middle 80% to the rest?
Added by Sheri D.
Step 1
Given: Mean (μ) = 170 Standard Deviation (σ) = 25 Calculate z-score for $140: \[ z = \frac{140 - 170}{25} = -1.2 \] Calculate z-score for $210: \[ z = \frac{210 - 170}{25} = 1.6 \] Show more…
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