Translate the following words to algebra symbols, using "n" as the variable. The product of six and a number is 24.

Andrew plans to retire in 32 years. He plans to invest part of his retirement funds in stocks, so he seeks out information on pa

Debora [2.8K]

Answer:

a) 0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.

b) 0.4129 = 41.29% probability that the mean return will be less than 8%

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.

Mean 8.7% and standard deviation 20.2%.

This means that $\mu = 8.7, \sigma = 20.2$

40 years:

This means that $n = 40, s = \frac{20.2}{\sqrt{40}}$

(a) What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 40 years will exceed 13%?

This is 1 subtracted by the pvalue of Z when X = 13. So

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

By the Central Limit Theorem

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

$Z = \frac{13 - 8.7}{\frac{20.2}{\sqrt{40}}}$

$Z = 1.35$

$Z = 1.35$ has a pvalue of 0.9115

1 - 0.9115 = 0.0885

0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.

(b) What is the probability that the mean return will be less than 8%?

This is the pvalue of Z when X = 8. So

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

$Z = \frac{8 - 8.7}{\frac{20.2}{\sqrt{40}}}$

$Z = -0.22$

$Z = -0.22$ has a pvalue of 0.4129

0.4129 = 41.29% probability that the mean return will be less than 8%

8 0

3 years ago

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

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Answer:

a) 0.5588

b) 0.9984

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 24

Standard Deviation, σ = 6.4

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(score between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{6.4} \leq z \leq \displaystyle\frac{30-24}{6.4}) = P(-0.62 \leq z \leq 0.94)\\\\= P(z \leq 0.94) - P(z < -0.62)\\= 0.8264 - 0.2676 = 0.5588 = 55.88\%$

$P(20 \leq x \leq 30) = 55.88\%$

b) Sampling distribution

Sample size, n = 22

The sample will follow a normal distribution with mean 24 and standard deviation,

$s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{6.4}{\sqrt{22}} =1.36$

c) P(mean score of sample is between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{1.36} \leq z \leq \displaystyle\frac{30-24}{1.36}) = P(-2.94 \leq z \leq 4.41)\\\\= P(z \leq 4.41) - P(z < -2.94)\\= 1 - 0.0016 = 0.9984 = 99.84\%$