The probability that a random sample of 100 dinners has a total weight less than 4760 hectograms is approximately 0.9772.
To find the probability that a random sample of 100 fried chicken dinners from the lot has a total weight less than 4760 hectograms, we can use the central limit theorem. The mean weight (μ) of a dinner is 48 hectograms, and the standard deviation (σ) is 2.4 hectograms.
The central limit theorem states that the distribution of the sample means approaches a normal distribution as the sample size increases, regardless of the shape of the original distribution. The mean of the sample means (μxˉ ) is equal to the population mean, and the standard deviation of the sample means (σxˉ) is calculated by dividing the population standard deviation by the square root of the sample size.
For this scenario, the mean of the sample means is 48 hectograms, and the standard deviation of the sample means is 0.24 hectograms.
Now, to find the probability of a total weight less than 4760 hectograms for a sample of 100 dinners, we convert this value to a z-score using the formula z= X−μxˉ/σxˉ. Substituting the values, we get z= 4760−48/0.24≈197.0.
Consulting a standard normal distribution table or using statistical software, we find that the probability (P(Z<197.0)) is extremely close to 1, or more precisely, approximately 0.9772. Therefore, there is a very high probability that a random sample of 100 dinners from the lot will have a total weight less than 4760 hectograms.