Answer:
e. 0.977
Step-by-step explanation:
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.
In this problem, we have that:

Which of the following is closest to the proportion of daily transactions greater than 350?
This is 1 subtracted by the pvalue of Z when X = 350. So



has a pvalue of 0.023
1 - 0.023 = 0.977
So the correct answer is:
e. 0.977
Answer:
4/7
Step-by-step explanation:
Since there are 7 possible outcomes because there are 7 triangles the denominator will be 7. Since there are 4 white squares the chances of landing on one is 4/7
Answer:
is 5
Step-by-step explanation:
I hope this helps you
X= 0.2034394 or 14.50799581 . I hope this helps
Answer:
a) Binomial distribution B(n=12,p=0.01)
b) P=0.007
c) P=0.999924
d) P=0.366
Step-by-step explanation:
a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.
b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

Then,

c) In this case, the distribution is B(1200,0.01)

d) In this case, the distribution is B(100,0.01)
We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.
