Mathematics, 03.04.2020 15:05 nicolemillan20

The weight of potato chip bags filled by a machine at a packaging plant is normally distributed with a mean of 15.0 ounces and a standard deviation of 0.1 ounces. What percentage of bags weigh more than 14.8 ounces

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Answer from: joelpimentel

97.73% of bags weigh more than 14.8 ounces.

Step-by-step explanation:

We are given that the weight of potato chip bags filled by a machine at a packaging plant is normally distributed with a mean of 15.0 ounces and a standard deviation of 0.1 ounces.

Let X = weight of potato chip bags filled by a machine

So, X ~ N( $\mu=15.0,\sigma^{2} =0.1^{2}$ )

The z-score probability distribution for normal distribution is given by;

Z = $\frac{ X -\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean weight = 15.0 ounces

$\sigma$ = standard deviation = 0.1 ounces

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, percentage of bags that weigh more than 14.8 ounces is given by = P(X > 14.8 ounces)

P(X > 14.8 ounces) = P( $\frac{ X -\mu}{\sigma}$ > $\frac{14.8-15.0}{0.1}$ ) = P(Z > -2) = P(Z < 2)

= 0.9773 or 97.73%

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.9773.

Hence, 97.73% of bags weigh more than 14.8 ounces.

Answer from: Quest

step-by-step explanation:

times by times

Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

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