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:
44 Give me brainliest please!
Step-by-step explanation:
180-90-46
90-46
44
Answer:
What's confusing ??? You literally just put a dot on the number line
Example:
1.3, move 3 units to the right of 1 and place a dot there
Step-by-step explanation:
By Remainder Theorem, f(5) = 91.
=> (5)³ - 4(5) + k = 91
=> 125 - 20 + k = 91
=> 105 + k = 91
=> k = -14.
Answer:
-125,-626
Step-by-step explanation:
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