CHAPTER 5 Distributions for 266 (b) Find the probability that a group of five di
ID: 3063000 • Letter: C
Question
CHAPTER 5 Distributions for 266 (b) Find the probability that a group of five digits from the table will contain at least one 0 (c) In Table B, there are 40 digits on any given line. What is the mean number of Os in lines 40 digits long? 5.30 Online learning. Recently, the U.S. Department of Education released a report on online learning stating that blended instruction, a combination of conventional face-to-face and online instruction, appears more effective in terms of student performance than conventional teaching.9 You decide to poll the incoming students at your institution to see if they prefer courses that blend face-to-face instruction with online components. In an SRS of 400 incoming students, you find that 311 prefer this type of course. (a) What is the sample proportion who prefer this type of blended instruction? (b) If the population proportion for all students nationwide is 85%, what is the standard deviation of p? (c) Using the 68-95-99.7 rule, if you had drawn an SRS from the United States, you would expect p to fall between what two percents about 95% of the time? (d) Based on your result in part (a), do you think that the incoming students at your institution prefer this type of instruction more, less, or about the same as students nationally? Explain your answer.Explanation / Answer
Question 53 .
N = total number of student = 400
n = number of student prefer to blended instructoion type of course = 311
a) Sample propertion who prefer blended type of instruction course = n /N = 311/400 = 0.7775
b) Population propertion for all students nationwide =P= 85%
S.D.(phat) = sqrt ( PQ /n ) where n = 400, P = 0.85 ,Q = 1-P = 0.15
S.D(phat) = 0.0178
c) Using 68- 95- 99.7
95% of estimate of phat lies in the interval ( E(phat) - 2 * S.D(phat) , E(phat) + 2 * S.D(phat) )
Where E(phat) =P = 0.85
S.D.(phat) = 0.0178
E(phat) - 2 * S.D(phat) = 0.85 - ( 2*0.0178) = 0.8144
E(phat) + 2 * S.D(phat) = 0.85+ ( 2*0.0178) = 0.8856
95% phat fall between ( 0.8144, 0.8856)
d) By using range of thumb
Minimum usual Value = E(phat) - 2 * S.D(phat) = 0.85 - ( 2*0.0178) = 0.8144
Maximum usual value = E(phat) + 2 * S.D(phat) = 0.85+ ( 2*0.0178) = 0.8856
Since p = 0.7775 < 0.8144
Hence Propertion of incoming student prefer to blended instruction type of course is significantly low.
Question 58.
p = probability of success = probability of getting 1 at each thrown = 1/6
Define the variable
X : Number of throws before getting 1 (i.e. number of failure before getting the first success)
Hence X ~ Geo(p =1/6)
and the pmf of random variable X is
P(X=x) = pqx = (1/6) (5/6)x ; x =0,1,2.........
a) Required Probability =P(X=1)
= (1/6) *(5/6)
= 0.1388
b) Required Probability = P(x=2)
= (1/6 ) * (5/6) 2
= 0.1157
c) i) Required probability = P(X=3)
= (1/6 ) * (5/6) 3
= 0.0964
ii) Required probability = P(X=4)
= (1/6 ) * (5/6) 4
= 0.0803