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A Level OCR Hard stats question

C718F0DB-3D07-4B9D-AE68-999BD75CAE84.jpegI have completed parts a,b and c. But how do i solve d??
Reply 1
Original post by Matheen1
C718F0DB-3D07-4B9D-AE68-999BD75CAE84.jpegI have completed parts a,b and c. But how do i solve d??

You can find the cumulative probability for the left tail
P(X<160)
and add 0.6 to it and take that away from 1. That would give the cumulative probability for the right tail which should be a calculator job to get the required value?

Note a sketch of a normal bell curve with the important values / areas marked on usually helps.
(edited 10 months ago)
Reply 2
Original post by mqb2766
You can find the cumulative probability for the left tail
P(X<160)
and add 0.6 to it and take that away from 1. That would give the cumulative probability for the right tail which should be a calculator job to get the required value?

How do I solve part d)
Reply 3
Original post by Matheen1
How do I solve part d)


How do I show that the model isn’t appropriate??
Reply 4
Original post by Matheen1
How do I solve part d)

Note the previous edit about sketching the normal / bell curve and marking important values/parameters on.

For d) what is the mean/std dev for the belll curve and roughly where would the min/max lie in relation to those (where on the sketch) so are they symmetric about the mean, roughly how many std dev away from the mean for 1000 data points?
(edited 10 months ago)
Reply 5
Original post by mqb2766
Note the previous edit about sketching the normal / bell curve and marking important values/parameters on.

For d) what is the mean/std dev for the belll curve and roughly where would the min/max lie in relation to those (where on the sketch) so are they symmetric about the mean, roughly how many std dev away from the mean for 1000 data points?

3F965B2E-03A4-4619-BE16-5A822543099E.jpeg
Can you just help me to understand the mark scheme for the answer?
Reply 6
Original post by Matheen1
3F965B2E-03A4-4619-BE16-5A822543099E.jpeg
Can you just help me to understand the mark scheme for the answer?


Under the assumptions of the model, what's the probability of a given rock being <112?

From the observed data, what's the probability of a given rock being < 112?

Do these probabilities match up?
(edited 10 months ago)
Reply 7
Original post by DFranklin
Under the assumptions of the model, what's the probability of a given rock being <112?

From the observed data, what's the probability of a given rock being < 112?

Do these probabilities match up?

Can you please show me how to do this as I think I’m doing something wrong. I am getting p=0.0712 but the markscheme says 0.0708.image.jpg
Reply 8
Original post by Matheen1
3F965B2E-03A4-4619-BE16-5A822543099E.jpeg
Can you just help me to understand the mark scheme for the answer?


You should know about the mean
+/-1 std dev corresponds to 68% or 32% in the tails, so for 1000 data points, you d have 320 outside +/-60
+/-2 std dev, corresponds to 95% or 5% in the tails, so for 1000 data points youd have ~50 lying outside +/-120
+/-3 std dev, corresponds to 99.7% or 0.3% in the tails, so for 1000 data points youd have ~3 lying (just) outside +/-180
so the rule of thumb would be that 88 (288-mean) must correspond to ~3 std devs.
(edited 10 months ago)
Reply 9
Original post by Matheen1
Can you please show me how to do this as I think I’m doing something wrong. I am getting p=0.0712 but the markscheme says 0.0708.image.jpg

Its in the noise, but agree with your value. their numbers correspond to 111.8
(edited 10 months ago)
Reply 10
Original post by mqb2766
Its in the noise, but agree with your value. their numbers correspond to 111.8


Ok so if the value of p(x<112) is greater than p(x<79.6) whcih is the lower limit of 2 standard deviations, then the model isn’t appropriate i.e. for the model to be appropriate p(x<112) should in theory be less than p(x<79.6)???
Reply 11
Original post by Matheen1
Can you please show me how to do this as I think I’m doing something wrong. I am getting p=0.0712 but the markscheme says 0.0708.

Looks like the markscheme is wrong, but the exact value doesn't matter.

According to the model the probability of a rock being < 112 is about 0.07

In 1000 rocks, we didn't see *any* rocks that small. If the model was correct, the chance of this happening is 0.93^1000 which is less than 1 in 10^30. Whether the exact value is 0.0712 or 0.07 isn't going to make any practical difference.
(edited 10 months ago)
Reply 12
Original post by Matheen1
Ok so if the value of p(x<112) is greater than p(x<79.6) whcih is the lower limit of 2 standard deviations, then the model isn’t appropriate i.e. for the model to be appropriate p(x<112) should in theory be less than p(x<79.6)???


For me, just thinking about the intervals so for 1000 data points (+/- 3 std dev) with mean=200 and std dev=60 youd expect the min/max values to be about [20,380]. The observed values are very different so a std dev of about 1/2 the models value is ~right. For +/- 3 std devs youd expect ~3 values just outside the interval so the min/max values would correspond to ~3.1 or 3.2 std devs from the mean.

The +/-2 std devs isnt the best comparison to make here, but its a common one in hypothesis testing so tails of 0.025 each (0.05 in total) and 0.95 in the main body between +/- 2 std devs so [80,320]. Youd expect 25 values less than the value of 80 out of 1000 and 80 is less than 112. The previous paragraph so +/- 3 std devs is the way to go.

Probabilistically, p(x<112)=0.07, which means youd expect 70 values out of the 1000 less than this value so the model aint great. Really youd want about 0.0005 in each tail and using that with std dev of 30 gives a min value of 102 which is about right, though the std dev value could be refined if really necessary.
(edited 10 months ago)
Reply 13
Original post by mqb2766
For me, just thinking about the intervals so for 1000 data points (+/- 3 std dev) with mean=200 and std dev=60 youd expect the min/max values to be about [20,380]. The observed values are very different so a std dev of about 1/2 the models value is ~right.

The +/-2 std devs isnt the best comparison to make here, but its a common one in hypothesis testing so tails of 0.025 each (0.05 in total) and 0.95 in the main body between +/- 2 std devs so [80,320]. Youd expect 25 values less than the value of 80 out of 1000 and 80 is less than 112. The previous paragraph so +/- 3 std devs is the way to go.

Probabilistically, p(x<112)=0.07, which means youd expect 70 values out of the 100 less than this value so the model aint great. Really youd want about 0.0005 in each tail and using that with std dev of 30 gives a min value of 102 which is about right, though the std dev value could be refined if really necessary.

So how can i use this to solve part e?
Reply 14
Original post by Matheen1
So how can i use this to solve part e?


Hopefully the previous two posts covered that? You want the min/max interval to correspond to +/-3 std devs, so what should the std dev (var) be?

Again, a sketch with imporant values/areas marked on usually helps me.
(edited 10 months ago)
Reply 15
Original post by Matheen1
Ok so if the value of p(x<112) is greater than p(x<79.6) whcih is the lower limit of 2 standard deviations, then the model isn’t appropriate i.e. for the model to be appropriate p(x<112) should in theory be less than p(x<79.6)???


I don't quite understand what you're trying to say here. It is surely obvious that p(x < 112) is always greater (or equal) than p(x<79.6), since if x < 79.6 then certainly x is also < 112.

The fundamental thing you need to be doing is comparing what the model says with what is seen in the observed data (and then making a logical argument about what this implies). If you're doing this anywhere, you're certainly not explaining it enough for me to follow.

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