Typing errors in a text are either nonword errors (as when \'the\' is typed as \
ID: 3149383 • Letter: T
Question
Typing errors in a text are either nonword errors (as when 'the' is typed as 'tch') 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. If X is the number of word errors missed, what is the distribution of X? X Is binomial with n = 15 and p = 0.7. X is binomial with n = 15 and p = 0.3. X is Normal with mu = 10.5 and alfa =- 1.77. X is approximately Normal with mu = 4.5 and alfa = 1.77. If Y Is the number of word errors caught, what is the distribution of Y? Y is Normal with mu = 10.5 and alfa = 1.77. y Is binomial with n = 15 and p = 0.7. Y is binomial with n = 15 and p = 0.3. Y is approximately Normal with mu = 4.5 and alfa = 1.77. What Is the mean number of errors caught? (Inter your answer to one decimal place.) What is the mean number of errors missed? (Enter your answer to one decimal place.) What is the standard deviation of the number of errors caught? (Round your answer to tour decimal places.) What is the standard deviation of the number of errors missed? (Round your answer to four decimal places.)Explanation / Answer
let p=probability of catching errors=70%=0.7
q=1-p=probability of missing the errors=0.3
n=15
since there are two out comes here are catching errors and missing errors, so it will follow binomial distribution
a) x=number of word error missed then its distribution will be binomial with n=15 and p=0.3
second choice is true
y=number of word error caught the its distribution will be binomial with n=15 and p=0.7
second choice is true
b) mean number of error caught=n*p=15*0.7=10.5
mean number of error missed=n*q=15*0.3=4.5
c) variance of mean number of error caught=n*p*(1-p)=15*0.7*0.3=3.15
standard deviation of mean number of error caught=sqrt(3.15)=1.7748
variance of mean number of error missed =n*q*(1-q)=15*0.3*0.7=3.15
standard deviation of mean number of error missed=sqrt(3.15)=1.7748
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