Evaluate the limit by first recognizing the sum as the Riemann sum for a function defined on the interval [0,1]: lim as n approa

ches infinity (1/n)((sqrt(1/n)+(sqrt(2/n)+(sqrt(3/n)+...............+(sqrt(n/n))

Mathematics

1 answer:

mina [271]3 years ago

3 0

$\lim_{n \to +\infty} \left ( \dfrac{1}{n} \cdot \left ( \sqrt{ \dfrac{1}{n} } + \sqrt{ \dfrac{2}{n} } + \sqrt{ \dfrac{3}{n} } + \cdots + \sqrt{ \dfrac{n}{n} }} \right ) \right ) =$

$\dfrac{1}{n} \sum_{i = 1}^n \left ( \sqrt{ \dfrac{i}{n} } \right ) = \int\limits^1_0 { \sqrt{x} } \, dx = \left [ \frac{2}{3}x^{ \frac{3}{2} } \right ]^{1}_{0} = \dfrac{2}{3} \cdot 1^{ \frac{3}{2} } - \dfrac{2}{3} \cdot 0^{ \frac{3}{2} } = \boxed{ \dfrac{2}{3} }$

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A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute

Katyanochek1 [597]

Answer:

Step-by-step explanation:

Given that:

sample size n = 36

standard deviation = 10.1

level of significance ∝ = 0.10

The null hypothesis and the alternative hypothesis can be computed as follows:

$H_0 : \sigma = 10$

$H_1 : \sigma \neq 10$

The test statistics can be computed as follows:

$X^2 = \dfrac{(n -1)s^2 }{\sigma ^2}$

$X^2 = \dfrac{(36 -1)10.1^2 }{10^2}$

$X^2 = \dfrac{(35)102.01 }{100}$

$X^2 = \dfrac{3570.35 }{100}$

$X^2 =35.704$

degree of freedom = n - 1 = 36 - 1 = 35

Since this test is two tailed .

The P -value can be determined by using the EXCEL FUNCTION ( = 2 × CHIDIST(35.7035, 35)

P - value = 2 × 0.435163515

P - value = 0.8703 ( to four decimal places)

Decision Rule : To reject the null hypothesis if P - value is less than the 0.10

Conclusion: We fail to reject null hypothesis ( accept null hypothesis) since p-value is greater than 0.10 and we conclude that there is sufficient claim that the normal range of pulse rates of adults given as 60 to 100 beats per minute resulted to a standard deviation of 10 beats per minute.