Software that can be used to detect fraund in customer
phonecards, tracks the number of metropolitan areas from which
callsoriginate each day. suppose 1% of the legitimate users
originatecalls from two or more metropolitan areas in a single day;
and 30%of fradulent users originate calls from two or more
metropolitanarea in a single day. the overall proportion of
fraudulent users is0.01%. if the same person originates calls from
two or moremetropolitan areas in as ingle, what is probability that
such useis fraudlent?

0.002991

Step-by-step explanation:

Let

User is legitimate one = E₁

User is fraudulent one = E₂

User originates calls from two metropolitan = A

Therefore, according to the question

P(E₁) = 1 - 0.01% = 1 - 0.0001 = 0.9999

P(E₂) = 0.0001

P(A/E₁) = 1% = 0.01

P(A/E₂) = 30% = 0.30

Now,

P(E₂/A) =

or

P(E₂/A) =

or

P(E₂/A) = 0.002991

here is your mark brainliest

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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