The amount of cereal in boxes, packed by a particular machine, is normally distributed with mean µ gram andstandard deviation 5 gram. If the advertised weight of a box is 500 gram, find(i) the proportion of boxes that will be underweight (i.e. weight less than 500 gram) when µ = 505;(ii) the value of µ required to ensure that only 1% of the boxes are underweight.(b) As a check on the setting of the machine a random sample of four boxes is chosen and the setting changed if theaverage weight of the four boxes is less than 500 gram. Find the probability that the setting of the machine ischanged when µ = 505.

(i) To find the proportion of boxes that will be underweight when µ = 505, we need to calculate the area under the normal distribution curve to the left of 500 grams.

First, we need to standardize the value of 500 grams using the formula z = (x - µ) / σ, where x is the value we want to standardize, µ is the mean, and σ is the standard deviation.

In this case, x = 500 grams, µ = 505 grams, and σ = 5 grams. Plugging these values into the formula, we get z = (500 - 505) / 5 = -1.

Next, we need to find the proportion of the area under the normal distribution curve to the left of z = -1. We can use a standard normal distribution table or a calculator to find this value.

Looking up the z-score of -1 in the standard normal distribution table, we find that the proportion of the area to the left of z = -1 is approximately 0.1587.

Therefore, the proportion of boxes that will be underweight when µ = 505 is approximately 0.1587 or 15.87%.

(ii) To find the value of µ required to ensure that only 1% of the boxes are underweight, we need to find the z-score corresponding to the desired proportion.

Using a standard normal distribution table or a calculator, we can find the z-score that corresponds to a proportion of 0.01 (1%).

Looking up the z-score of -2.33 in the standard normal distribution table, we find that the proportion of the area to the left of z = -2.33 is approximately 0.01.

Now, we can use the formula z = (x - µ) / σ to solve for µ. Rearranging the formula, we get µ = x - z * σ.

Plugging in the values of x = 500 grams, z = -2.33, and σ = 5 grams, we can calculate µ as follows: µ = 500 - (-2.33) * 5 = 511.65 grams.

Therefore, to ensure that only 1% of the boxes are underweight, the value of µ should be approximately 511.65 grams.

(b) To find the probability that the setting of the machine is changed when µ = 505, we need to calculate the probability that the average weight of a random sample of four boxes is less than 500 grams.

Since the sample size is small (n = 4), we need to use the t-distribution instead of the normal distribution.

Using the t-distribution, we can calculate the t-score using the formula t = (x̄ - µ) / (s / √n), where x̄ is the sample mean, µ is the population mean, s is the sample standard deviation, and n is the sample size.

In this case, x̄ is the average weight of the four boxes, µ = 500 grams, s is the standard deviation of the population (5 grams), and n = 4.

Let's assume that the average weight of the four boxes is x̄ = 498 grams. Plugging in these values, we get t = (498 - 500) / (5 / √4) = -0.8.

Now, we can use a t-distribution table or a calculator to find the probability that the t-score is less than -0.8.

The probability that the t-score is less than -0.8 is approximately 0.2257.

Therefore, the probability that the setting of the machine is changed when µ = 505 is approximately 0.2257 or 22.57%.