Answer:
B) (4,5)
Step-by-step explanation:
Hope this helps.. (again)
suppose the people have weights that are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Find the probability that if a person is randomly selected, his weight will be greater than 174 pounds?
Assume that weights of people are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Mean = 177
standard deviation = 26
We find z-score using given mean and standard deviation
z = 
= 
=-0.11538
Probability (z>-0.11538) = 1 - 0.4562 (use normal distribution table)
= 0.5438
P(weight will be greater than 174 lb) = 0.5438
Answer:
41.6 cm
Step-by-step explanation:
Answer:
5 negative numbers
Step-by-step explanation:
-5 -4 -3 -2 -1 0 1 2 3 4 5
1 2 3 4 5 6 7 8 9 10 11
The variance of the binomial distribution of the number of households with landline service is 2.
<h3>What is the binomial probability distribution?</h3>
It is the probability of <u>exactly x successes on n repeated trials, with p probability</u> of a success on each trial.
The variance of the distribution is given by:
V(X) = np(1 - p)
In this problem, the parameters are given as follows:
n = 8, p = 0.504.
Hence the variance is given by:
V(X) = 8 x 0.504 x 0.496 = 2.
More can be learned about the binomial distribution at brainly.com/question/24863377
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