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katovenus [111]
3 years ago
5

The pulse rates of 174 randomly selected adult males vary from a low of 39 bpm to a high of 111 bpm. Find the minimum sample siz

e required to estimate the mean pulse rate of adult males. Assume that we want 90% confidence that the sample mean is within 3 bpm of the population mean. Complete parts​ (a) through​ (c) below. a. Find the sample size using the range rule of thumb to estimate σ.
Mathematics
1 answer:
Daniel [21]3 years ago
8 0

Answer:

a. The sample size is approximately 97 adults

Step-by-step explanation:

The given parameters are;

The number of adults surveyed, n = 174

The lowest value of the pulse rate = 39 bpm

The highest value of the pulse rate = 111 bpm

The level of confidence used for determining the sample size = 90%

The given sample mean = 3 pm of the population mean

a. The range rule of thumb states that the standard deviation is approximately one quarter (1/4) of the range

The range = 111 bpm - 39 bpm = 72 bpm

Therefore, the standard deviation, σ = 72 bpm/4 = 18 bpm

The sample size, 'N', is given as follows;

N = \dfrac{z^2 \cdot p \cdot q}{e^2}

Where;

N = The sample size

z = The confidence level, 90% (z-score at 90% = 1.645)

p·q = σ² = 18² = 324

e² = 3² = 9

N = \dfrac{1.645^2  \times 324}{3^2} = 97.4169

Therefore the appropriate sample size, N ≈ 97 adults.

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(i) Given that

V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr

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In Mathematica, you can first define the velocity function with

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Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr

Compute the integral:

V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr

V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr

V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R

V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)

V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

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