Question

In: Math

A) Fill in the blanks. According to estimates by the office of the Treasury Inspector General...

A) Fill in the blanks. According to estimates by the office of the Treasury Inspector General of IRS, approximately 0.07 of the tax returns filed are fraudulent or will contain errors that are purposely made to cheat the IRS. In a random sample of 332 independent returns from this year, around __________ returns, give or take __________, will be fraudulent or will contain errors that are purposely made to cheat the IRS.

1)

23.24 , 21.60

2)

332 , 4.649

3)

23.24 , 0.07

4)

23.24 , 4.649

5)

4.649 , 23.24

B)

According to estimates by the office of the Treasury Inspector General of IRS, approximately 0.0354 of the tax returns filed are fraudulent or will contain errors that are purposely made to cheat the IRS. In a random sample of 393 independent returns from this year, what is the probability that greater than 10 will be fraudulent or will contain errors that are purposely made to cheat the IRS?

1)

0.8232

2)

0.0337

3)

0.8907

4)

0.1768

5)

0.0675

C)

Suppose that the probability of a baseball player getting a hit in an at-bat is 0.2884. If the player has 40 at-bats during a week, what's the probability that he gets exactly 16 hits?

1)

0.0205

2)

0.0409

3)

0.0451

4)

0.9591

5)

0.9549

Solutions

Expert Solution

a)
n = 332 , p = 0.07

Here, μ = n*p = 23.24, σ = sqrt(np(1-p)) = 4.649

b)

Here, n = 393, p = 0.0354, (1 - p) = 0.9646 and x = 9
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >10).
P(X <= 9) = (393C0 * 0.0354^0 * 0.9646^393) + (393C1 * 0.0354^1 * 0.9646^392) + (393C2 * 0.0354^2 * 0.9646^391) + (393C3 * 0.0354^3 * 0.9646^390) + (393C4 * 0.0354^4 * 0.9646^389) + (393C5 * 0.0354^5 * 0.9646^388) + (393C6 * 0.0354^6 * 0.9646^387) + (393C7 * 0.0354^7 * 0.9646^386) + (393C8 * 0.0354^8 * 0.9646^385) + (393C9 * 0.0354^9 * 0.9646^384)
P(X <= 9) = 0 + 0 + 0.0001 + 0.0004 + 0.0013 + 0.0036 + 0.0085 + 0.0172 + 0.0305 + 0.0479
P(X <= 9) = 0.1095


P(x> 10) = 1- P(x< =9)
= 1 - 0.1095
= 0.8907

c)

Here, n = 40, p = 0.2884, (1 - p) = 0.7116 and x = 16
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 16)

P(x = 16) = 40C16 * 0.2884^16 *(1-0.2884)^24
= 0.0409


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