Please help with this question.The mean weight of the contents of samples of 30 bags of sugar has
standard error 0.008kg.The mean is 1.0042kg.Standard deviation is 0.042kg Choose the option that is closest to the
probability, to three decimal places, that the mean weight of the
contents of samples of 30 bags of sugar will be 1kg or more.
Options for Question
A 0.700 B 0.800 C 0.824
D 0.858 E 0.887 F 0.932
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
To find the probability that the mean weight of the contents of samples of 30 bags of sugar will be 1kg or more, we can use the standard error and the normal distribution.
The standard error measures the variability of the sample means around the true population mean. In this case, the standard error is 0.008kg.
We are given that the mean weight is 1.0042kg. To calculate the z-score, which represents the number standard deviations away from the mean, we subtract the desired mean (1kg) from the given mean and divide by the standard error:
z = (1 - 1.0042) / 0.008 = -0.525
Now, we need to find the probability that the z-score is less than -0.525. By looking up the z-score in a standard normal distribution table or using a calculator, we can find the corresponding probability.
Based on the options given, we can see that option C (0.824) is the closest to the probability. However, the calculated value may differ slightly depending on the level of precision used in the z-score calculation.