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alexgriva [62]
3 years ago
6

You’re given 3 timings, all in the format of mm:ss (29:12 for example would mean 29 minutes and 12 seconds). They are labelled a

s x, y and z.
When they’re all rounded to the nearest whole number, x, y and z are 29, 29 and 13 respectively, all in minutes.

Thus, when x, y and z are added together, what are the chances that the result is 72 minutes, when rounded to the nearest whole number.
Mathematics
1 answer:
katrin2010 [14]3 years ago
5 0
29 + 29 + 13 = 71
The range of possible times in seconds for x is 28:30 - 29:29. Similarly for y.
The range of possible times for z is 12:30 - 13:29
The range of possible values for x + y + z = 69:30 - 72:27 which is 178 seconds long
To be recorded as 72 minutes the sum would be in the range 71:30 - 72:27 which is 58 seconds long
P(result recorded as 72 min) = 58/178

If you simply used the ranges [28.5, 29.5) [28.5, 29.5) and [12.5, 13.5)
then the sum would be in the range [69.5, 72.5)
and P(result recorded as 72 min) = 60 seconds / 180 seconds = 1/3

Using discrete times in seconds gives a slightly different answer but is also much harder to understand.
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Answer:

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Step-by-step explanation:

The sample size is n=64.

We start by calculating the sample mean and standard deviation with the following formulas:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}

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The sample standard deviation is s=6.89.

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{6.89}{\sqrt{64}}=\dfrac{6.89}{8}=0.861

The degrees of freedom for this sample size are:

df=n-1=64-1=63

The t-value for a 99% confidence interval and 63 degrees of freedom is t=2.656.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.656 \cdot 0.861=2.29

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 21.52-2.29=19.23\\\\UL=M+t \cdot s_M = 21.52+2.29=23.81

The 99% confidence interval for the mean is (19.23, 23.81).

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Answer:

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