I need an answer fast please! The first digits of data entries in most real-world data sets are not uniformly distributed. The most common first digit is 1, followed by 2, and so on, with 9 being the least common first digit. This phenomenon is known as Benford's Law. The table below is the probability distribution for Benford's Law where the random variable DDD represents the first digit in a data entry. DDD P(D)P(D)P, left parenthesis, D, right parenthesis DDD P(D)P(D)P, left parenthesis, D, right parenthesis 111 0.3010.3010, point, 301 666 0.0670.0670, point, 067 222 0.1760.1760, point, 176 777 0.0580.0580, point, 058 333 0.1250.1250, point, 125 888 0.0510.0510, point, 051 444 0.0970.0970, point, 097 999 0.0460.0460, point, 046 555 0.0790.0790, point, 079 Find the indicated probability. P(D \leq 2)=P(D≤2)=P, left parenthesis, D, is less than or equal to, 2, right parenthesis, equals

0.699

Step-by-step explanation:

P(D \geq 2)P(D≥2)P, left parenthesis, D, is greater than or equal to, 2, right parenthesis is the probability that the first digit is 222, 333, 444, and so on, all the way up to 999.

Hint #22 / 3

Let's see how likely it is for the first digit to be 222 or higher:

DDDP(D)P(D)P, left parenthesis, D, right parenthesisDDDP(D)P(D)P, left parenthesis, D, right parenthesis

111{0.301}0.3010, point, 301666\blueD{0.067}0.067start color #11accd, 0, point, 067, end color #11accd

222\blueD{0.176}0.176start color #11accd, 0, point, 176, end color #11accd777\blueD{0.058}0.058start color #11accd, 0, point, 058, end color #11accd

333\blueD{0.125}0.125start color #11accd, 0, point, 125, end color #11accd888\blueD{0.051}0.051start color #11accd, 0, point, 051, end color #11accd

444\blueD{0.097}0.097start color #11accd, 0, point, 097, end color #11accd999\blueD{0.046}0.046start color #11accd, 0, point, 046, end color #11accd

555\blueD{0.079}0.079start color #11accd, 0, point, 079, end color #11accd  

We could add those probabilities to find our answer, but it's probably easier to use the fact that all of the probabilities add to 111, then subtract the probabilities we don't want:

P(D \geq 2)=1-0.301=0.699P(D≥2)=1−0.301=0.699P, left parenthesis, D, is greater than or equal to, 2, right parenthesis, equals, 1, minus, 0, point, 301, equals, 0, point, 699

Hint #33 / 3

The

P(D \geq 2)=0.699P(D≥2)=0.699

0.699

Step-by-step explanation:

I did it on khan

step-by-step explanation:

a.

step-by-step explanation:

step 1 distribute 2 to every term in the parentheses

step 2 add 10 on both sides of the equal sign