\"Durable press\" cotton fabrics are treated to improve their recovery from wrin
ID: 3327958 • Letter: #
Question
"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. Unfortunately, the treatment also reduces the strength of the fabric. The breaking strength of untreated fabric is normally distributed with mean 51.8 pounds and standard deviation 2.3 pounds. The same type of fabric after treatment has normally distributed breaking strength with mean 23.5 pounds and standard deviation 2 pounds. A clothing manufacturer tests 3 specimens of each fabric. All 6 strength measurements are independent. (Round your answers to four decimal places.) (a) What is the probability that the mean breaking strength of the 3 untreated specimens exceeds 50 pounds? (b) What is the probability that the mean breaking strength of the 3 untreated specimens is at least 25 pounds greater than the mean strength of the 3 treated specimens? eBook Submit Answer Save Progress Practice Another Version + 1 points MIntroStat95.E.517.XP. My Notes Ask Your Teacher 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 10 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 eBookExplanation / Answer
1 Solution:-
a) P(Z > (50 - 51.8)/(2.3/sqrt(3))) = P(Z > -1.3555) = P(Z < 1.3555 ) = 0.9131
b) Because the untreated and treated sample means are independent as well, the difference of the untreated minus treated sample means is normally distributed with mean 51.8 - 23.5 = 28.3 pounds and standard deviation sqrt[(2.3/sqrt(3))^2 + (1.9/sqrt(3))^2] = 1.7224 pounds
P(difference of the untreated minus treated sample means >= 25)
= P(Z >= (25 - 28.3)/1.7224
= P(Z >= -1.9159)
= 0.9726
2 solution:-
P(X>m) = 0.05
where m is the misses
n = 10, p = 0.3
so average/expected number of misses = 10 *0.3 = 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(m) >= 0.95
we wil calculate value of each and then when the sum
0.0282 + 0.1210 + 0.2335 + 0.2668 + 0.2001 + 0.1029 + 0.0368 + 0.0090
= 0.9983
we have sufficent evidence that a proofreader atually cathces fewer than 70% of word errors
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.