The manager of the commercial mortgage department of a large bank has collected data during the past two years concerning the nu
1 answer:
The <em>expected number of mortgages</em> approved per week and the standard deviation of the distribution are 2.019 and 0.024 respectively.
<u>The expected number of mortgages approved per week</u> :
- <em>Mean = (Σfx ÷ Σf)</em>
Expected Number approved = 210 ÷ 104 = 2.019
Hence, it is expected that 2.019 mortageahes would be approved per week.
<u>The standard deviation</u> :
- <em>Variance = [Σ(Xi - x)² ÷ Σf] </em>
- <em>Standard deviation = √Variance</em>
Variance = (59.5414 ÷ 104) = 0.0005698
Standard deviation = √0.0005698
Standard deviation = 0.024
Therefore, the expected value and standard deviation are 2.019 and 0.024 respectively.
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For the remaining limit, use the Taylor expansion for cos(x) :
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Then
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Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.
Edit: some scratch work suggests the limit is 10 for n = 6.