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
Hello :) we got similar answers but i believe you made a few distributing mistakes! hopefully my work makes sense, if not i’ll totally try to explain more
p.s i might be too late whoops
p.p.s i’m in a higher grade but i haven’t done this kinda math in a longgg time haha so if you think i did something wrong trust your gut ✌️
Answer:
GIVE ME MY RAMEN
Step-by-step explanation:
3.
add like terms
30 + 7k = 100
100-30 = 70
70 divided by 7k = 10
k = 10
4. add the z's together
2z - 6 -2 = -10
+ 6 + 2
-10 will be -2 now since we dragged the -6 and -2 to the other side
2z = -2
z = -1
5. 3.2x - 1.7x = 1.5x
1.5x + 5.5 = 10
10 would be 4.5 since we dragged 5.5 to the other side so it would be 10 - 5.5
3.2x = 5.5
x = 1.71875
6.
3/4x - 1/4x = 2/4x
14 - 3 = 11
2/4x = 11
x = 22
Answer:
the answere for this question is (b) because it has equal differences
