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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
The image is missing, so i have attached it
Answer:
Equation is;
q/sin 105 = 15/sin 30
Step-by-step explanation:
We are given;
Angle Q = 105°
Angle S = 30°
s = 15
From sine rule we know that;
a/sin A = b/sin B = c/sin C.
Thus, applying that to this question, we can find side q.
Thus;. q/Sin Q = s/Sin S
Plugging in the relevant values to obtain ;
q/sin 105 = 15/sin 30
