Answer:
its blurry for me sorry
Step-by-step explanation:
I feel like it would be 5/12, but im not 100% sure
Answer:
C. unlikely
Step-by-step explanation:
Problems of normally distributed samples can be 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.
A probability is said to be extremely likely if it is 95% or higher, and extremely unlikely if it is 5% or lower. A probabilty higher than 50% and lower than 95% is said to be likely, and higher than 5% and lower than 50% is said to be unlikely.
In this problem, we have that:

How likely is it that a single survey would return a mean of 30%?
We have to find the pvalue of Z when X = 0.30.



has a pvalue of 0.1587.
So the correct answer is:
C. unlikely
Answer:
3
Step-by-step explanation:
dwdwdwada
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.