Asoft 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.

The probability of filling a cup between 30 and 31 ounces 9.2%

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

x is the raw score, μ is mean and σ is standard deviation

Given that μ = 28, σ = 2, hence:

From the normal distribution table, P(30 < x < 31) = P(1 < z < 1.5) = P(z < 1.5) - P(z < 1) = 0.9332 - 0.8413 = 9.2%

The probability of filling a cup between 30 and 31 ounces 9.2%

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Step-by-step

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)

expression #3 is not equivalent. the others are.

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3