A soft drink machine outputs a mean of 28 ounces per cup. The machine's output is normally distributed with a standard deviation of 2 ounces. What is the probability of filling a cup between 30 and 31 ounces? Round your answer to four decimal places.

Normally distributed => get ready the normal probability table, or equivalent.

mu = mean = 28 oz

sigma = standard deviation = 2 oz

Need to find P(30 to 31),  the probability of filling between 30 and 31 oz.

Solution:

With normal probabilities,

P(30 to 31) = P(X<31) - P(X<30), i.e. the difference of the right tails for X=30 and X=31.

Calculate the Z scores

Z(X) = (X-mu)/sigma

Z(31) = (31-28)/2 = 1.5

P(X<31) = P(Z<1.5) = 0.9331928

Z(30) = (30-28)/2 = 1.0

P(X<30) = P(Z<1.0) = 0.8413447

Therefore,

probability of filling between 30 and 31 oz

= P(X<31) - P(X<30)

= 0.9331928 -  0.8413447

= 0.09184805

= 0.0918 (to 4 decimal places)

Step-by-step explanation:

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)

i do not know the answer

step-by-step explanation: