Question

Use technology or a z-distribution table to find the indicated area. The weights of boxes of rice produced at a factory are normally distributed with a mean of 24 oz and a standard deviation of 1.3 oz. Consider a shipment of 1200 boxes of rice. How many of the boxes will weigh 25 oz or less?

Answer 1

$\mathbb P(X\le25)=\mathbb P\left(\dfrac{X-24}{1.3}\le\dfrac{25-24}{1.3}\right)\approx\mathbb P(Z\le0.7692)$

Consulting a z-table, you should find a probability of about $0.7791=77.91\%$.

So of the total 1200 boxes, you should expect 77.91% of them to weigh less than 25 oz, or approximately 935.

Answer 2

The expected number of the boxes that will weigh 25 oz or less from the 1200 boxes of rice in shipment is 935.

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

$X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$ )

then it can be converted to standard normal distribution as

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

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

$P(Z \leq z) = P(Z < z) )$

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

$P(Z \leq z) = \rm p \: value$

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

The probability that out of n trials, there'd be x successes is given by

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

The expected value and variance of X are:

$E(X) = np\\ Var(X) = np(1-p)$

Let we take:

X = the weights of the rice boxes, in oz, produced in the considered factory.

Then, as per the problem's statement, we get;

$X \sim N(\mu = 24, \sigma = 1.3)$

The probability that X gets value 25 or less is:

$P(X \leq 25)$

Converting X to standard normal distribution, we get:

$Z = \dfrac{X - \mu}{\sigma} = \dfrac{X - 24}{1.3}$

The probability  $P(X \leq 25)$ can be rewritten as:

$P(X \leq 25) = P\left( Z \leq \dfrac{25 - 24}{1.3} \right) \approx P(Z \leq 0.77)$

From the z-tables, the p-value for Z = 0.77 is 0.7794

Thus, we get:

$P(X \leq 25) \approx P(Z \leq 0.77) = 0.7794$

or

$P(X \leq 25) \approx 77.94\%$

Let we take:

n = 1200 boxes being bernoulli experiments, each of them prone to success(having weight less or equal to 25 oz) with probability p = 0.7794 or failure (weight > 25 oz) with probability q = 1-p = 0.2206.

And, Y = the number of successes in those 1200 trials.

Then we get:

$Y \sim B(n = 1200, p = 0.7794)$

The expected value of Y will be the expected number of successes, so it would be the expected number of the boxes that will weigh 25 oz or less.

We get:
$E(Y) = np = 1200 * 0.7794 \approx 935 \: \rm (in \: integer)$

Thus, the expected number of the boxes that will weigh 25 oz or less from the 1200 boxes of rice in shipment is 935

Learn more about binomial distribution here:

https://brainly.com/question/13609688

Learn more about z-score here:

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Learn more about standard normal distribution here:

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