Need help answering question 7.62 :/ please help. (d) P( X = 1.5). (2) E(X) = 4.

Question

Need help answering question 7.62 :/ please help.

(d) P( X = 1.5). (2) E(X) = 4. (x) f) Var(= o . 61 In , a random variableX, is 7each given case, defined. You are to a binomial distribution. Justify your answer by referring t are not satisfied . You (a) Six days a week for a yearyou play the three-digit win on a given day if your three-digit number matc State of Michigan. For the first six monthsyou select the number last six months, you select the number 234 each day. Let X be t" the year that your choice is a winner. (b) Studies have found that 80% of dorm residents get along with their ton 0 room in a certain dorm has two occupants. A survey was conduct at random and asking each occupant "Do you get along with your rol the number of 'yes's” among the 20 responses obtained. decide whether or not X he specific assumptions that are or criek Three Michigan hours . the one selected at random by t for the , 123 each day, number of days in Each : by rooms ?” Let X be A 7.An experiment is test 62 designed to whether a subject has ESP, extrasensory perception. total of 96 cards is drawn, one by one, with replacementfrom a well-shuffled ordinary as suit of each card drawn. are of cards. The subject is asked to guess the There four possible. choices for each card—namely, spades, hearts, diamonds, and clubs. We wish to test the nu

Explanation / Answer

a)

If you guess, the probability of getting it right is 1/4 = 0.25.

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   <=   0.25
Ha:   p   >   0.25 [ANSWER]

****************************

b)

We want to get the probability of "at least 35" correct out of 96.

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    96      
p = the probability of a success =    0.25      
x = our critical value of successes =    35      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   34   ) =    0.991675702
      
[This part is binomcdf(96,0.25,34).]
  
Thus, the probability of at least   35   successes is  
          
P(at least   35   ) =    0.008324298

Hence,

Pvalue = 0.008324298 [ANSWER]

*********************************

c)

As P < 0.05, we reject Ho.

****************************

d)

There is significant evidence at 0.05 level that the subject has ESP, that is, the probability of him guessing right is more than 0.25.

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.