Im attempting the following question with this provided information.
μ = 675 σ = 21
A gardener receives a shipment of 30 bags of soil, find the probability that the total weight is less than 20kg. Explain why the information that the weight of the bags follow a normal distribution is not needed to answer this question.
Originally I attempted the following:
- For the total weight of the shipment to be less than 20kg than the average weight of the bags must be less than the total weight (g) over the number of bags. 20000g / 30 = 667g (3 dp.)
With 667g being the sample mean for the shipment we can find a z-score to find out if the total weight of the shipment will be below 20kg. z = (667 - 675) / (21 / sqrt(30)) = -2.09
Probability from |z| = 0.0183
So my answer was that there is a 1.83% chance the shipment will be below 20kg.
But since the question states that we can answer the question without the information of the weight of bags following a normal distribution i'm inclined to think that using the populations st.dev to calculate my answer is an incorrect way to go about the question.
Is there another method I can use the calculate this question, otherwise what could possibly be meant that we do not need the normal distribution information to answer the question.