A) Daily sales of petrol from the Nabua service station are normally distributed with

Mean6300Landthe standard deviation 400L.
(i) If a daily sale is selected at random, find the probability that it is less than 6200L.
(ii) If petrol sales are sampled for 40 days, and the mean is calculated, find the

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i)

P(X < 6200)
P(Z < 6200-6300/400)
P(Z < - 0.25) = 0.4013

The probably of sales being less than 6200L is 40.13%

ii) Question is not complete, so I cannot answer it.

To find the probability and mean in this scenario, we can use the properties of the normal distribution.

(i) To find the probability that a daily sale is less than 6200L, we need to calculate the area under the normal curve up to the value of 6200L.

Step 1: Standardize the value of 6200L.
We use the formula:
Z = (X - μ) / σ
where Z is the standardized score, X is the value we want to standardize, μ is the mean, and σ is the standard deviation.

In this case, X = 6200L, μ = 6300L, and σ = 400L.
Plugging in these values, we get:
Z = (6200 - 6300) / 400

Step 2: Look up the standardized score in the Z-table.
The Z-table provides the area to the left of a given Z-score. In this case, we want to find the area to the left of the standardized score we calculated in Step 1.

Assuming we get a standardized score of -0.25, we can look up this value in the Z-table to find the corresponding area. This area represents the probability.

(ii) If petrol sales are sampled for 40 days, and the mean is calculated, we need to calculate the mean of the sample means.

The mean of the sample means is equal to the population mean, which is 6300L, in this case.

Please note that to calculate the mean of the sample means, we assume that each sample mean is calculated from random and independent samples, following the same distribution as the population.

By following these steps, you can find the probability and mean in the given scenario related to the petrol sales data.