Asoft-drink machine can be regulated so that it discharges an average of μ oz. per cup. if the ounces of fill are normally distributed with a standard deviation of 0.4 oz., what value should μ be set at so that 98% of 6-oz. cups will not overflow?

μ = 5.068 oz

Step-by-step explanation:

Normal distribution formula to use the table attached

Z = (x - μ)/σ

where μ is mean, σ is standard deviation, Z is on x-axis and x is a desired point.

98% of 6-oz. cups will not overflow means that the area below the curve is equal to 0.49; note that the curve is symmetrical respect zero, so, 98% of the cases relied between the interval (μ - some value) and (μ + some value)].

From table attached, area = 0.49 when Z = 2.33. From data, σ = 0.4 oz and x = 6 oz (maximum capacity of the cup). Isolating x from the formula gives

Z = (x - μ)/σ

2.33 = (6 - μ)/0.4

μ = 6 - 2.33*0.4

μ = 5.068

This means that with a mean of 5 oz and a standard deviation of 0.4 oz, the machine will discharge a maximum of 6 oz in the 98% of the cases.

50x5=$250

step-by-step explanation:

b. hoped i'm right lol

step-by-step explanation: