The mean weight (1.0042kg)of the contents of samples of 30 bags of sugar has
standard error 0.008kg. 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
To find the probability that the mean weight of the contents of samples of 30 bags of sugar will be 1kg or more, we need to use the concept of the standard error and the normal distribution.
First, we need to calculate the z-score, which measures the number of standard errors the desired value is away from the mean. In this case, the desired value is 1kg, and the mean is 1.0042kg. The formula for calculating the z-score is:
z = (x - μ) / SE
Where:
x = desired value
μ = mean weight
SE = standard error
Substituting the given values:
z = (1 - 1.0042) / 0.008
Simplifying further:
z = -0.0042 / 0.008
z ≈ -0.525
Next, we need to find the cumulative probability (also known as the area under the curve) for the z-score. We can use a standard normal distribution table or a calculator to look up the closest value to -0.525.
Looking up the z-score in the standard normal distribution table, we find that the closest value is approximately 0.300.
The probability that the mean weight of the contents of samples of 30 bags of sugar will be 1kg or more is equal to 1 minus the cumulative probability (1 - 0.300):
P(mean weight ≥ 1kg) = 1 - 0.300
P(mean weight ≥ 1kg) ≈ 0.700
Therefore, the option closest to the probability, to three decimal places, is:
A) 0.700