There are two brands of Corn Flakes, Post and Kellogs. Each brand has the same size box. However, because of each brand’s filling procedure, they have different mean weights. The weights of a box of Post Corn Flakes is approximately normal with μ = 64.1 oz and σ = .5 oz while the weight of a box of Kellogs, which is also normally distributed, has μ = 63.9 oz and σ = .4 oz.

A box is selected from each brand and weighed. What is the probability that the Post box will outweigh the Kellogs box?

Probability of an event is the measure of its chance of occurrence. The probability that the post box will outweigh the Kellogs box is 0.4129 approximately.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z-score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] standard deviation [tex]\sigma[/tex])

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

[tex]P(Z \leq z) = P(Z < z) )[/tex]

Also, know that if we look for Z = z in z-tables, the p-value we get is

[tex]P(Z \leq z) = \rm p \: value[/tex]

What is the distribution of random variable which is sum of normal distributions?

Suppose that a random variable X is formed by [tex]n[/tex] mutually independent and normally distributed random variables such that:

[tex]X_i = N(\mu_i , \sigma^2_i) ; \: i = 1,2, \cdots, n[/tex]

And if

[tex]X = X_1 + X_2 + \cdots + X_n[/tex]

Then, its distribution is given as:

[tex]X \sim N(\mu_1 + \mu_2 + \cdots + \mu_n, \: \: \sigma^2_1 + \sigma^2_2 + \cdots + \sigma^2_n)[/tex]

If, for the given case, we assume two normally distributed random variables as:

X = variable assuming weights of boxes of Post Corn Flakes
Y = variable assuming weights of boxes of Kellogs

Then, as per the given data, we get:

[tex]X \sim N(\mu = 64.1, \sigma = 0.5)\\Y \sim N(\mu = 63.9, \sigma = 0.4)[/tex]

Then, the probability that the Post box will outweigh the Kellogs box can be written as:

[tex]P(X > Y)[/tex]

Or,

[tex]P(X -Y > 0)[/tex]

We need to know about the properties of X-Y.

Also, since [tex]E(aX) = aE(X), Var(aX) = a^2Var(X)[/tex], thus,

[tex]-Y \sim N(-63.9, 0.4)[/tex]

As both are independent(assuming), thus,

[tex]X - Y \sim N(\mu = 64.1 - 63.9, \sigma = 0.5 + 0.4) = N(0.2, 0.9)[/tex]

Using the standard normal distribution, we get the needed probability as:

[tex]P(X -Y > 0) = 1 - P(X - Y \leq 0) \\P(X -Y > 0)= 1- P(Z = \dfrac{(X-Y) - \mu}{\sigma} \leq \dfrac{0 - 0.2}{0.9})\\P(X -Y > 0) \approx 1 - P(Z \leq -0.22)[/tex]

Using the z-tables, the p-value for Z = -0.22 is 0.4129

Thus, [tex]P(X > Y) = P(X - Y > 0) \approx 0.4129[/tex]

Thus, the probability that the post box will outweigh the Kellogs box is 0.4129 approximately.

Learn more about standard normal distribution here:

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