Question 1198237: A magazine article states that in a study of 64 students, average teenager reads 5.5 books a year with a standard deviation of 2 books. What is the probability that the average teenager actually reads less than 5 books?
One of the following is the correct answer. Which one?
A) 0.0228 (2.28%)
B) 0.9772 (97%)
C) 0.5987 (59.87%)
D) 0.4013 (40.13%(
Found 2 solutions by ewatrrr, ikleyn:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! Normal Distribution (continuous curve) µ = 5.5 and σ = 2
probability that the average teenager actually reads less than 5 books?
Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
P(x<5) = normcdf(-9999,5,2,5.5) = .4013
0r
area to the 'left' of z = (5-5.5_/2 = -.25, P(z<.25) = .4013
Would include reading 4.5 books, for example
IF the variable x < 5 needed to be a whole number:
Then, p(x < 5 ) = p( x ≤ 4) = .2266 would work
Answer by ikleyn(52810)  (Show Source):
You can put this solution on YOUR website! .
As the problem is presented in this post, the problem is INCORRECT.
The solution by @ewatrrr is INCORRECT, too.
To learn WHY, read my solution below to the end.
" . . . less than 5 books " means that the problem, ACTUALLY, asks
+--------------------------------------------------------+
| What is the probability that the average teenager |
| actually reads 4 or less books ? |
+--------------------------------------------------------+
Therefore, the correct formula expression is
P = normalcfd(-9999, 4, 5.5, 2)
and it is equal to 0.2266. ANSWER
Solved.
The correct answer in this problem is P = 0.2266 (rounded).
It is not in the list - - - therefore, the problem is presented incorrectly in the post.
@ewatrrr calculated P = normalcfd(-9999, 5, 5.5, 2) instead of P = normalcfd(-9999, 4, 5.5, 2), giving incorrect answer.
In joking form, regarding this problem,
I would say that it is a double trap.
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