Step 1 :Given that the weights of broilers are normally distributed with a mean of 1423 grams and a standard deviation of 169 grams.
Step 2 :We are asked to find the 33rd percentile, 90th percentile, third quartile of the weights, and the weight that the chicken farmer should guarantee so that he will have to give his customers' money back only 1% of the time.
Step 3 :We can use the formula for the z-score, which is \(X = Z*σ + μ\), where X is the data point, μ is the mean, and σ is the standard deviation. The Z value corresponds to the percentile and can be found using a standard normal distribution table or a calculator with statistical functions.
Step 4 :Using this formula, we find that the 33rd percentile of the weights is approximately \(1348.65\) grams.
Step 5 :The 90th percentile of the weights is approximately \(1639.58\) grams.
Step 6 :The third quartile of the weights is approximately \(1536.99\) grams.
Step 7 :The chicken farmer should guarantee a weight of at least \(1029.85\) grams to only have to give his customers' money back 1% of the time.
Step 8 :Final Answer: \(\boxed{1348.65}\) grams for the 33rd percentile, \(\boxed{1639.58}\) grams for the 90th percentile, \(\boxed{1536.99}\) grams for the third quartile, and \(\boxed{1029.85}\) grams for the weight guarantee.