A soft-drink machine can be regulated so 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 6-02 cups will overflow only 2% of the time?

u = 5.176 oz

Step-by-step explanation:

Given:

standard deviation = 0.4

P (x>6) = 2℅ = 0.02

The complement role:

P (A°) = P (not A)

= 1 - P (A)

P (x6) = 1 = 0.02 =

= 1-0.02 = 0.98

Taking the closest probability 0.9893 which is on row 2.0 column. 06 in the probability table, we now have a z score of

2.0 + 0.06 = 2.06

Zo = 2.06

Therefore, Z score is expressed as

Zo = (x - u) / sd

We already know Zo = 2.06

x = 6

u = ?

sd = 0.4

Therefore,

2.06 = (6 - u) / 0.4

= 2.06 * 0.4 = 6 - u

u = 5.176

The correct answer is µ = 5.178

Step-by-step explanation:

Find µ for P (X > 6) = 0.02

Normal distribution

σ = 0.4 oz

P (X > Xo) = 0.02 = > P (X < Xo) = 0.98

P (X - µ/σ < Xo - µ/σ) = 0.98

P (Z < Xo - µ / σ) = 0.98

Xo - µ / σ = invNorm (0.98)

µ = Xo - µ x invNorm (0.98)

Use the z-table to get value for invNorm (0.98) which is 2.054

µ = 6=0.4 x (2,054) = 5178400 or 5.178

µ = 5.178