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
Hi,
16,500 rounded to the nearest ten thousand is 20,000.
Have a great day!
A. 1/2 ÷ 1/4= 1/2 x 4/1 = 4/2 = 2
b 4/5÷2/3 = 4/5 x 3/2 = 12/10= 6/5= 1 1/5
c. 7 ÷ 1/4 = 7x 4/1 = 28/1 = 28
d. 5/12÷ 5/6 = 5/12 x 6/5 = 30/60 = 1/2
e 3 1/3 ÷3/4 = 10/3 x 4/3 = 40/9= 4 4/9
f. 18/25 ÷ 3/5= 18/25 x 5/3 = 90/75 = 18/15= 1 3/15= 1 1/5
g. 1 ÷ 9/13 = 1/1 x 13/9= 13/9 = 1 4/9
h 1 3/5 ÷5/6 = 8/5 x 6/5 = 48/25 = 1 23/25
I 7/8÷7/16= 7/8x 16/7= 112/56 = 2
Answer:
c
Step-by-step explanation:
tye both have the same amount on each side