Typographic errors in a text are either nonword errors (as when \"the\" is typed

Question

Typographic 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 proofreaders catch 70% of word errors. You ask a fellow student to proofread an essay in which you have deliberately made 20 word errors. What is the smallest number of misses m with P(X m) no larger than 0.05? You might consider m or more misses as evidence that a proofreader actually catches fewer than 70% of word errors. _____misses

Explanation / Answer

Modeling this as a Binomial distribution with the following parameters:

n = 20, p = 0.70

We have to calculate the value of 'm' such that:

P(X >= m) < 0.05

So,

1 - P(X >= m) > 0.95

Also,

1 - P(X >= m) = P(X<m)

So,

P(X < m) > 0.95

Now,

P(X < m) = P(X=0) + P(X=1) + P(X=2) +...+ P(X=m-1)

P(X=0) = nC0*p0*(1-p)n-0 = 20C0*0.70*0.320 = 3.48*10-11

P(X=1) = 20C1*0.71*0.319 = 1.62*10-9

P(X=2) = 20C2*0.72*0.318 = 3.6*10-8

P(X=3) = 20C3*0.73*0.317 = 3.6*10-8

Similarly,

P(X=7) = 0.001

P(X=8) = 0.0038

P(X=9) = 0.012

P(X=10) = 0.03

P(X=11) = 0.065

P(X=12) = 0.114

P(X=13) = 0.164

P(X=14) = 0.191

P(X=15) = 0.179

P(X=16) = 0.13

Upto P(X=6), values are negligible, so we add from P(X=7) upto this point.

P(X=0) +...+ P(X=16) = P(X=7) +...+ P(X=16) = 0.001+0.0038+0.012+0.03+0.065+0.114+0.164+0.191+0.179+0.13 = 0.89

So, we can go further.

P(X=17) = 0.07

Now,

P(X=0) +...+ P(X=17) = 0.89+0.07 = 0.96

At this point we have exceeded the value 0.95

So, we have:

m = 17

Hope this helps !

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