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Montano1993 [528]
2 years ago
8

19. A sample of 50 retirees is drawn at random from a normal population whose mean age and standard deviation are 75 and 6 years

, respectively.19a. Describe the shape of the sampling distribution of the sample mean in this case.19b. Find the mean and standard error of the sampling distribution of the sample mean.19c. What is the probability that the mean age exceeds 73 years
Mathematics
1 answer:
Juliette [100K]2 years ago
5 0

Answer:

a) Approximately normal.

b) The mean is 75 years and the standard deviation is 0.8485 years.

c) 0.9909 = 99.09% probability that the mean age exceeds 73 years

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Sample of 50 retirees

This means that n = 50

Mean age and standard deviation are 75 and 6 years

This means that \mu = 75, \sigma = 6

a. Describe the shape of the sampling distribution of the sample mean in this case

By the Central Limit Theorem, approximately normal.

b. Find the mean and standard error of the sampling distribution of the sample mean.

By the Central Limit Theorem, the mean is 75 and the standard error is s = \frac{6}{\sqrt{50}} = 0.8485

c. What is the probability that the mean age exceeds 73 years?

This is 1 subtracted by the pvalue of Z when X = 73.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{73 - 75}{0.8485}

Z = -2.36

Z = -2.36 has a pvalue of 0.0091

1 - 0.0091 = 0.9909

0.9909 = 99.09% probability that the mean age exceeds 73 years

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Use the distributive property

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Offer B is the best value for money because each kg cost is £0.61 and in offer 'A' each kg cost is £0.7.

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We have:
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12 + 5 + 32 × 10 − 5 = <br> (12 + 5) + 32 × (10 − 5) =
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What are the solutions to the system of equations? {y=2x2+6x−10 y=−x+5
inessss [21]
Answer: x = -5, 3/2 y = 10, 7/2

Explanation:

y = 2x^2 + 6x - 10 (1)
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=> 2x^2 + 6x - 10 = - x + 5
2x^2 - 7x - 15 = 0
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According to (2):

y = -x + 5
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f(x) = 3 cos(x) 0 ≤ x ≤ 3π/4 evaluate the Riemann sum with n = 6, taking the sample points to be left endpoints. (Round your ans
Kruka [31]

Answer:

\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558

Step-by-step explanation:

We want to find the Riemann sum for \int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx with n = 6, using left endpoints.

The Left Riemann Sum uses the left endpoints of a sub-interval:

\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)

where \Delta{x}=\frac{b-a}{n}.

Step 1: Find \Delta{x}

We have that a=0, b=\frac{3\pi }{4}, n=6

Therefore, \Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}

Step 2: Divide the interval \left[0,\frac{3 \pi}{4}\right] into n = 6 sub-intervals of length \Delta{x}=\frac{\pi}{8}

a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b

Step 3: Evaluate the function at the left endpoints

f\left(x_{0}\right)=f(a)=f\left(0\right)=3=3

f\left(x_{1}\right)=f\left(\frac{\pi}{8}\right)=3 \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}=2.77163859753386

f\left(x_{2}\right)=f\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}=2.12132034355964

f\left(x_{3}\right)=f\left(\frac{3 \pi}{8}\right)=3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=1.14805029709527

f\left(x_{4}\right)=f\left(\frac{\pi}{2}\right)=0=0

f\left(x_{5}\right)=f\left(\frac{5 \pi}{8}\right)=- 3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=-1.14805029709527

Step 4: Apply the Left Riemann Sum formula

\frac{\pi}{8}(3+2.77163859753386+2.12132034355964+1.14805029709527+0-1.14805029709527)=3.09955772805315

\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558

5 0
2 years ago
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