CFA Practice Question
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CFA Practice Question
The bolts produced by M and K Steel have lengths with a mean of 3 cm and a standard deviation of 0.3 cm. A box of 200 of these bolts is considered a random sample. What is the probability that the average lengths of the bolts in the box are shorter than 2.95 cm?
Correct Answer: B
Because the sample size is large (n > 30) the distribution of the x-bars is approximately normal with mean 3 and standard deviation 0.3/sqrt(200) = 0.02. Now, the z-score for 2.95 is (2.95 - 3)/0.02 = -2.5. The table value for -2.5 is 0.0062. So, P(x-bar < 2.95) = 0.62%. The probability that the average length of the bolts in the box of 200 bolts is less than 2.95 cm is 0.62%.
User Contributed Comments 15
|minna||how did you calculate, vincent?|
|gambary||how are you supposed to know the probability without a z table on the exam??|
|ehc0791||depending on how many digits you take on
0.3/sqrt(200), if you take 4, it will be 0.0212
then you will get the final result of .9%
|gkobylko||Remember that for normal distribuition the probabilities of values falling within a number of deviations are:
1 stdev = 68%
1.65 stdev = 90%
1.96 stdev = 95%
2.5 stdev = 99%
the z-score for this question is 2.357, close to 2.5, so the answer will be close to (1-.99)/2=.5%
Only one obvious answer, no need to know the whole z table, just the key values
|DannyZhou||What is 0+ and 1-?|
|tschorsch||some very tiny amount>0 (i.e. like 0.000001217372)
some very tiny amount<1 (i.e. like 0.999991231231)
|UFCRP||gkobylko, are you replying to Gambary? How would you discern 0.62% from normal distribution to get the correct answer. My guess is, you would need a Z table, wouldn't you?
|DonAnd||vincent i got the same answer as you using all decimal places. I guess it's best to see how much d.p. is given in the amswer|
|tankdan||should provided z-table for these problems|
|johntan1979||Best thing is don't round the intermediate steps. 0.94% is the precise answer.|
|GouldenOne||How is this question any different than the college X an Y example? They use 2 different methods to solve this problem
|sgossett86||this is b*llsh*t. -.05/.02121 = 2.357 so what the hell are they getting a z of 2.5 for?|
|sgossett86||wow i feel like a retard. confidence intervals.. tail is half of the remainder to one. .5 on the south tail on the 99% CI. 2.5 on the south on the 95%. .6>.5 so just north of the 2.68 or < 2.68.
I took a two week break from studying and look where it got me... jeeze...
I need to hammer again. No more girlfriend...
|maryprz14||LOVE LOVE LOVE LOVEEEEE ANALYST NOTES...|