1. 2. Typing errors in a text are either nonword errors (as when \"the\" is type

Question

1.

2.

Typing errors in a text are either nonword errors (as when "the" is typed as 'teh") or word errors that result in a real but incorrect word. Spell-checking software will catch nonword errors but not word errors. Human of word errors, You ask a fellow student to proofread an essay in which you have deliberately made 12 word errors ers catch 70% (a) If the student matches the usual 70% rate·what is the distribution of the number of errors caught? What is the distribution of the number of errors missed? (b) Missing three or more out of 12 errors seems a poor performance. what is the probability that a proofreader who catches 70% of word errors misses exactly three out of 12? (Round your answer to four decimal places.) What is the probability of missing three or more out of 12? (Round your answer to four decimal places.) eBook

Explanation / Answer

a)
n = 12 , p =0.7

n = 12 , p = 0.3

b)
By binomial distribution,
As per binomial distribution,
P(X=r) = nCr * p^r * (1-p)^(n-r)

Here, n =12 , p = 0.3

P(x =3) = 12C3 * 0.3^3 * (1 -0.3)^9
= 0.2397

P(x <3) = P(X = 0) + P(X = 1) + P(X=2)

P(X>=3) = 1 - P(x <3)
= 0.7472

Excel formula to calculate this: =1-BINOM.DIST(2,12,0.3,TRUE)

2)
By binomial distribution,
As per binomial distribution,
P(X=r) = nCr * p^r * (1-p)^(n-r)

Here, n =17 , p = 0.3

a)P(x = 3)= 17C3 * 0.3^3 * (1 -0.3)^14
= 0.1245

b)
p(X <=3) = p(x =0) + P(x =1) + p(x =2) + P(x=3)
= 17C0 * 0.3^0 * (1 -0.3)^17 +17C1 * 0.3^1 * (1 -0.3)^16 +17C2 * 0.3^2 * (1 -0.3)^15 + 17C3 * 0.3^3 * (1 -0.3)^14
= 0.0023+0.0169+0.0581+0.1245
= 0.2018

c)
P(X >=3) = 1 - P(X<3)
= 1 - (0.0023+0.0169+0.0581)
=0.9227

d)
P(X<3) = p(x =0) + P(x =1) + p(x =2)
= 17C0 * 0.3^0 * (1 -0.3)^17 +17C1 * 0.3^1 * (1 -0.3)^16 +17C2 * 0.3^2 * (1 -0.3)^15
= 0.0773

e)
P(X>3) = 1 - p(X<=3)
= 1 - (0.0023+0.0169+0.0581+0.1245)
= 0.7982

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