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:
"23,520"
Step-by-step explanation:
one trip: 14 passengers
one day: 3 trips
passengers: 80
80x14= 1,120
1,120x3= 3,360
3,360x7= 23,520
Answer:
x=-2
Step-by-step explanation:
add 5x , then subtract 45, then divide by 15
Answer:
cuz i am batman hhhhhhhhhhhh