Final answer:
To find the level l for a 1% chance that the average weight of a SRS of 5 chocolate bars is less than l, we need the population mean and standard deviation and would use the z-score for a 1% chance in the formula involving the standard normal distribution.
Step-by-step explanation:
The question is asking to find the value of l such that there is a 1% chance that the average weight of a simple random sample (SRS) of 5 chocolate bars is less than l. To calculate this, we need to know the population mean (\(μ\)) and standard deviation (\(σ\)) of the weight of chocolate bars. Then, we can use the characteristics of the normal distribution, assuming the underlying distribution of weights is normal or approximately normal given a sufficiently large sample size.
First, we must standardize the problem referencing a standard normal distribution and use the z-score associated with a 1% chance (which is, typically, a z-score around -2.33 for a left-tailed test). We would apply the following formula for determining the sample mean threshold:
\(l = μ + (z × (σ / √ n))\)
where \(z\) is the z-score, \(n\) is the sample size (in this case, 5), \(μ\) is the population mean, and \(σ\) is the population standard deviation.
Without the actual values of \(μ\) and \(σ\), we cannot calculate the exact level of \(l\). However, this is the strategy one would apply to find the level \(l\) given the parameters of the normal distribution representing the chocolate bar weights.