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
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Answer:
11 units.
Step-by-step explanation:
Answer:
1/428 or 0.00233644859% chance
Step-by-step explanation:
Simply because there is only one page 45 use 428 as your denominator and then divide 1 by 428!
There is not an answer to an expression.
-3+p is an expression, unless there was an equal sign and a sum at the end, that is when we can solve for the variable x.
-3+p simply equals -3+p
