The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 415 grams and a standard deviation of 23 grams. Find the weight that corresponds to each event.

(a) The highest 20% weight correspond to the weight 434.32 grams.

(b) The middle 60% weight correspond to the weights 434.32 grams and 395.68 grams.

(c) The highest 80% weight correspond to the weight 395.68 grams.

(d) The highest 80% weight correspond to the weight 391.08 grams.

Step-by-step explanation:

Let X = weight of a small Starbucks coffee.

It is provided:

(a)

Compute the value of x fro P (X > x) = 0.20 as follows:

Use a standard normal table.

The value of z is 0.84.

The value of x is:

Thus, the highest 20% weight correspond to the weight 434.32 grams.

(b)

Compute the value of x fro P (x₁ < X < x₂) = 0.60 as follows:

Use a standard normal table.

The value of z is 0.84.

The value of x₁ and x₂ are:

Thus, the middle 60% weight correspond to the weights 434.32 grams and 395.68 grams.

(c)

Compute the value of x fro P (X > x) = 0.80 as follows:

Use a standard normal table.

The value of z is -0.84.

The value of x is:

Thus, the highest 80% weight correspond to the weight 395.68 grams.

(d)

Compute the value of x fro P (X < x) = 0.15 as follows:

Use a standard normal table.

The value of z is -1.04.

The value of x is:

Thus, the highest 80% weight correspond to the weight 391.08 grams.

5.6%

step-by-step explanation:

10 * 563.67 = 5636.7