Answer:
Step-by-step explanation:
Given that the weights of bags filled by a machine are normally distributed with a standard deviation of 0.055 kilograms and a mean that can be set by the operator.
Let the mean be M.
Only 1% of the bags weigh less than 10.5 kilograms
i.e. P(X<10.5) = 0.01
corresponding Z value for P(Z<z) = 0.01 is -0.025
i.e. 10.5 = M-0.025(0.055)
Solve for M from the above equation
M =
Rounding off we get
10.50 kgs
Mean weight should be fixed as 10.50 kg.
Answer:
0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?
This is 1 subtracted by the pvalue of Z when X = 3.1. So
By the Central Limit Theorem
has a pvalue of 1.
1 - 1 = 0
0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.