Here is a simple probability model for multiple-choice tests. Suppose that each
ID: 3327842 • Letter: H
Question
Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher p than a weak student.) The correctness of answers to different questions are independent. Jodi is a good student for whom p = 0.84.
(a) Use the Normal approximation to find the probability that Jodi scores 78% or lower on a 100-question test. (Round your answer to four decimal places.
(b) If the test contains 250 questions, what is the probability that Jodi will score 78% or lower? (Use the normal approximation. Round your answer to four decimal places.)
(c) How many questions must the test contain in order to reduce the standard deviation of Jodi's proportion of correct answers to half its value for a 100-item test?
questions
(d) Laura is a weaker student for whom p = 0.79. Does the answer you gave in (c) for standard deviation of Jodi's score apply to Laura's standard deviation also? Yes, the smaller p for Laura has no effect on the relationship between the number of questions and the standard deviation. No, the smaller p for Laura alters the relationship between the number of questions and the standard deviation.
"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 52.1 pounds and standard deviation 2.5 pounds. The same type of fabric after treatment has normally distributed breaking strength with mean 28.3 pounds and standard deviation 1.8 pounds. A clothing manufacturer tests 4specimens of each fabric. All 8 strength measurements are independent. (Round your answers to four decimal places.)
(a) What is the probability that the mean breaking strength of the 4 untreated specimens exceeds 50 pounds?
(b) What is the probability that the mean breaking strength of the 4 untreated specimens is at least 25 pounds greater than the mean strength of the 4 treated specimens?
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 12 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
a.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.84
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.84*0.16/100)
=0.0367
P(X > 0.78) = (0.78-0.84)/0.0367
= -0.06/0.0367 = -1.6349
= P ( Z >-1.635) From Standard Normal Table
= 0.949
P(X < = 0.78) = (1 - P(X > 0.78)
= 1 - 0.949 = 0.051
b.
P(X > 0.78) = (0.78-0.84)/0.0232
= -0.06/0.0232 = -2.5862
= P ( Z >-2.586) From Standard Normal Table
= 0.9951
P(X < = 0.78) = (1 - P(X > 0.78)
= 1 - 0.9951 = 0.0049
c.
since standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.78*0.22/100)
=0.041424
Since std = sqrt(pq/n)
(1/2)sqrt(pq/n) = sqrt(pq/4n)
0.0427/2 = sqrt ( 0.78*0.22 ) / 4n
0.02135 = sqrt ( 0.78*0.22 ) / 4n
0.0004558225 = ( 0.78 * 0.22 ) / 4n
4n = ( 0.78 * 0.22 / 0.0004558225
4n = 376.4623
n = 94.115 ~ 95
d.
Yes, the smaller p for Laura has no effect on the relationship between the
number of questions and the standard deviatio