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 15 word errors. What is the smallest number of misses m with P(X greaterthanorequalto 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.

Explanation / Answer

Here P (X >=m) <= 0.05

where m is the misses

so average/expected number of misses = 15 * 0.3 = 4.5

n = 15, p = 0.3

but we can't take normal approximation as np< 10

so we wil solve it by binomial.

Here we have to check that

P (X >=m) <= 0.05

so P( X<m) >= 0.95

so P(0) + P(1) + P(2) + P(3) +....P(m) >= 0.95

we wil calculate value of each and then when the sum reaches above 0.95, we will come to know

P(0) = 15C0 (0.3)0 ( 0.7) 15 = 0.0047

Total probability sum = 0.0047

P(1) = 15C1 (0.3)1 ( 0.7) 14 = 0.0305

Total probability sum = 0.0047 + 0.0305 = 0.0352

P(2) = 15C2 (0.3)2 ( 0.7) 13= 0.0916

Total probability sum = 0.0916 + 0.0352 = 0.1268

P(3) = 15C3 (0.3)3 ( 0.7) 12= 0.1700

Total probability sum = 0.1268 + 0.1700 = 0.2968

P(4) = 15C4(0.3)4 ( 0.7) 11= 0.2186

Total probability sum = 0.2186+ 0.2968 = 0.5154

P(5) = 15C5(0.3)5 ( 0.7) 10= 0.2061

Total probability sum = 0.5154+ 0.2061 = 0.7215

P(6) = 15C6(0.3)6 ( 0.7)9= 0.1472

Total probability sum = 0.7215 + 0.1472 = 0.8687

P(7) = 15C7(0.3)7 ( 0.7) 8= 0.0811

Total probability sum = 0.8687 + 0.0811 = 0.9498

so it is approximately 0.95 so m = 9 at which or above it , we have sufficent evidence that a proofreader actually cathces fewer than 70% of word errors.

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