Answer to Question #343357 in Statistics and Probability for Dzreke Rejoice

Question

Assume that the daily S&P return follows the normal distribution with mean μ = 0.00032 and standard deviation a= 0.00859.

a. Find the 75th percentile of this distribution.

b. Find the probability that the daily S & P return will be larger than 0 01

C. Consider the sample average S & P of a random sample of 20 days.

27:2 i. Describe the distribution of the sample mean with its expected value and standard error.

ii. What sample size is necessary to double the standard error of the mean?

iii. Is it more likely that the average S & P will be greater than 0 007, or that one day's S & P return will be?

iv. Find the number b such that P(x > b) = 0975

Answer 1

27.1

a. The formula below is used to compute percentiles of a normal distribution.

$$x=\mu+z\sigma$$

Using $Z=0.6745$ the 75^th percentile is

$$x=0.00032+0.6745(0.00859)=0.006114$$

b.

$$P(X>0.01)=1-P(Z\le\dfrac{0.01-0.00032}{0.00859})$$

$$\approx1-P(Z\le1.12689)\approx0.129894$$

c.

$$\mu_{\bar{X}}=\mu=0.00032$$

27.2

i.

$$\bar{X}\sim N(\mu, \sigma^2/n)$$

$n=20$

$$\bar{X}\sim N(0.00032, (0.00859)^2/20)$$

ii.

The larger the sample size, the smaller the standard error because the statistic will approach the actual value.

We have to decrease the sample size in 4 times to double the standard error of the mean.

$$n_2=5$$

iii.

$$P(X>0.007)=1-P(Z\le\dfrac{0.007-0.00032}{0.00859})$$

$$\approx1-P(Z\le0.777648)\approx0.218388$$

$$P(\bar{X}>0.007)=1-P(Z\le\dfrac{0.007-0.00032}{0.00859/\sqrt{20}})$$

$$\approx0.00025$$

One day's return is more likely to be greater than $0.007.$

iv.

$$P(X>b)=0.975$$

$$1-P(Z\le\dfrac{b-0.00032}{0.00859})=0.975$$

$$\dfrac{b-0.00032}{0.00859}=-1.96$$

$$b=0.00032-0.00859(1.96)$$

$$b=-0.0165$$