A sample survey interviews an SRS of 209 college women. Suppose that 69% of all college women have been on a diet within the last 12 months. What is the probability that 72% or more of the women in the sample have been on a diet

0.0869

Step-by-step explanation:

Given data:

No of college women = 209

Probability that 72% had been on a diet

Solution:

Standard deviation = √ p ( 1-p)/n

= √ 0.69 ( 1-0.69) / 209

= 0.02213

Z = x - u / standard deviation

= 0.72 - 0.69 / 0.02213

= 1.36

P( p ≥ 72 ) =( z< -13.56 )

= 0.0869

The probability that 72% of the women has been exercising for 12 month is 0.0869.

found it xd it ok yt

10.416 m/s

step-by-step explantion:

distance=100m

time= 9.6s

average speed=100/9.6

=10.416 m/s