The local supermarket buys lettuce each day to ensure really fresh produce. Each morning any lettuce that is left from the previous day is sold to a dealer that resells it to farmers who use it to feed their animals. This week the supermarket can buy fresh lettuce for $4.00 a box. The lettuce is sold for $10.00 a box and the dealer that sells old lettuce is willing to pay $1.50 a box. Past history says that tomorrow's demand for lettuce averages 250 boxes with a standard deviation of 34 boxes. R

Given Information: Cost price = $4Selling price = $10Salvage value = $1.50Average demand = μ = 250 boxes Standard deviation = σ = 34 boxes Required Information: Number of lettuce boxes = ?Answer: The supermarket should purchase 268 boxes  of lettuceStep-by-step explanation: The required number of lettuce boxes that supermarket should purchase is given by Number of lettuce boxes = μ + (z*σ) Where μ is the average demand of lettuce boxes, σ is the standard deviation, and z is the z-score which is given byp = C_us/(C_us + C_os) The z-score corresponding to probability p will be obtained. The cost of under stocking is given by C_us = Selling price - Cost price C_us = $10 - $4C_us = $6The cost of over stocking is given by C_os = Cost price - Salvage value C_os = $4 - $1.50C_os = $2.50 p = C_us/(C_us + C_os) p = 6/(6 + 2.50) p = 0.7058p  = 70.58%The z-score corresponding to 70.58% probability is approximately 0.54Number of lettuce boxes = μ + (z*σ) Number of lettuce boxes = 250 + (0.54*34) Number of lettuce boxes = 250 + (18.36) Number of lettuce boxes = 268.36Number of lettuce boxes ≈ 268Therefore, the supermarket should purchase 268 boxes of lettuce.How to use z-table? Step 1: In the z-table, find the probability you are looking for and note down the two-digit number of the given row. (e.g 0.5, 2.2, 0.5 etc.) Step 2: Then look up at the top of z-table and note down the value in the column for the remaining decimal point in the range of 0.00 to 0.09.Step 3: Finally, add the numbers obtained from step 1 and step 2.