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# Stats Question HELP! watch

1. The lengths of videos of a certain popular song have a normal distribution with mean 3.9 minutes. 18% of these videos last for longer than 4.2 minutes.
(i) Find the standard deviation of the lengths of these videos.
(ii) Find the probability that the length of a randomly chosen video differs from the mean by less than half a minute.
The lengths of videos of another popular song have a normal distribution with the same mean of
3.9 minutes but the standard deviation is twice the standard deviation in part (i)The probability that the length of a randomly chosen video of this song differs from the mean by less than half a minute is
denoted by p.
(iii) Without any further calculation, determine whether p is more than, equal to, or less than your answer to part (ii). You must explain your reasoning.

Sol. (i) I got s.d = 0.328
I am really confused and found diffcult on how to do on other two parts. Please help me out.
The lengths of videos of a certain popular song have a normal distribution with mean 3.9 minutes. 18% of these videos last for longer than 4.2 minutes.
(i) Find the standard deviation of the lengths of these videos.
(ii) Find the probability that the length of a randomly chosen video differs from the mean by less than half a minute.
The lengths of videos of another popular song have a normal distribution with the same mean of
3.9 minutes but the standard deviation is twice the standard deviation in part (i)The probability that the length of a randomly chosen video of this song differs from the mean by less than half a minute is
denoted by p.
(iii) Without any further calculation, determine whether p is more than, equal to, or less than your answer to part (ii). You must explain your reasoning.

Sol. (i) I got s.d = 0.328
I am really confused and found diffcult on how to do on other two parts. Please help me out.
Let's say X is the length of a video then X ~ N(3.9, 0.328).

If a video differs from the mean by less than a minute then you need the time to be between 2.9 and 4.9.

So you need to find P(2.9 < X < 4.9). Can you find that?

EDIT : Read the question wrong.
3. ...and without wishing to talk over Notnek, I wonder if I can also suggest sketching a bell curve and shading the two tails that fall outside of the specification? If you can find the area of (say) the upper tail, you can assume by symmetry that the lower tail will have the same area.

For part (iii), you could draw a new bell curve that is wider but lower (so as still to have a total area under the curve equal to 1). But the tails still start at the same absolute values, so...
4. (Original post by old_engineer)
...and without wishing to talk over Notnek, I wonder if I can also suggest sketching a bell curve and shading the two tails that fall outside of the specification? If you can find the area of (say) the upper tail, you can assume by symmetry that the lower tail will have the same area.
I was going to suggest the same thing but then I remembered these new calculators can work out P(a < X < b) just by setting the lower and upper limits. So I was going to leave it to the OP to have a go and see what method they used.
5. (Original post by old_engineer)
...and without wishing to talk over Notnek, I wonder if I can also suggest sketching a bell curve and shading the two tails that fall outside of the specification? If you can find the area of (say) the upper tail, you can assume by symmetry that the lower tail will have the same area.

For part (iii), you could draw a new bell curve that is wider but lower (so as still to have a total area under the curve equal to 1). But the tails still start at the same absolute values, so...
Oh okay I will try that too. Thanks
6. (Original post by Notnek)
Let's say X is the length of a video then X ~ N(3.9, 0.328).

If a video differs from the mean by less than a minute then you need the time to be between 2.9 and 4.9.

So you need to find P(2.9 < X < 4.9). Can you find that?
Btw how you got 2.9 and 4.9 because it says it differs from the mean by less than a minute. As i checked the marking scheme, it says 4.4 and 3.5 as half a minute means 0.5 less than 3.9. Please check the marking scheme and make me more clear according to it from the beginning. It would help a lot since my exams are near. Thank you
Q.5. (I hope the attachment works)
Attached Images
7. 9709_s17_ms_62.pdf (332.4 KB, 114 views)
Btw how you got 2.9 and 4.9 because it says it differs from the mean by less than a minute. As i checked the marking scheme, it says 4.4 and 3.5 as half a minute means 0.5 less than 3.9. Please check the marking scheme and make me more clear according to it from the beginning. It would help a lot since my exams are near. Thank you
Q.5. (I hope the attachment works)

Find the probability that the length of a randomly chosen video differs from the mean by less than half a minute.

I thought it said less than a minute. So if the mean is 3.9 then the region you're interested in has bounds 3.9 + 0.5 = 4.4 and 3.9 - 0.5 = 3.4. So you need to find

P(3.4 < X < 4.4)

Since the region is symmetrical around the mean, the above can be found by considering the regions to the left and right of the required region, which have equal probability.

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