Ina certain county weights of women are normally distributed with a mean of 138 lb and a standard deviation of 15 lb. What percentage of women in that country weigh more than 120 lb

88.493%

Step-by-step explanation:

We solve this question using z score formula.

Z score = x - μ/σ

where

x is the raw score = 120lb

μ is the population mean = 138lb

σ is the population standard deviation = 15lb

Hence,

z = 120 - 138/15

= -1.2

Probability value from Z-Table:

P(x<120) = 0.11507

P(x>120) = 1 - P(x<120)

1 - 0.11507

= 0.88493

Converting to percentage

= 0.88493 × 100

= 88.493%

The percentage of women in that country weigh more than 120 lb is 88.493%

answer: i think it's a

step-by-step explanation:

1,150 give brainliest

step-by-step explanation

2r= 2300

r =1150