Question

A certain stock market had a mean return of 2.6% in a recent year. Assume that the returns for stocks on the market were distributed normally, with a mean of 2.6 and a standard deviation of 10. Complete parts (a) through (g) below.

a. If you select an individual stock from this population, what is the probability that it would have a return less than 0 (that is, a loss)?

The probability is (Round to four decimal places)

b. If you select an individual stock from this population, what is the probability that it would have a return between -11 and -19?

The probability is (Round to four decimal places)

c. If you select an individual stock from this population, what is the probability that it would have a return greater than -7?

The probability is (Round to four decimal places)

d. If you select a random sample of four stocks from this population, what is the probability that the sample would have a mean return less than 0 (a loss)?

e. If you select a random sample of four stocks from this population, what is the probability that the sample would have a mean return between -11 and -19?

The probability is (Round to four decimal places)

Answer 1

Let X denotes the return from a randomly selected stock.

X ~ Normal(2.6, 10²)

a)  If you select an individual stock from this population, the probability that it would have a return less than 0 (that is, a loss)

- P(X<0)

X-2.6 0 2.6 P( 10 10

P(Z-0.26)

P(0.26)

0.397432

b) The probability that it would have a return between -11 and -19

= P(19X < -11)

= P(X 11) - P(X-19)

-11 2.6-P(2.6 -19 2.6 X- 2.6 PC 10 10 10 10 VI

= P(Z 1.36) P(Z 2.16)

= P(1.36) (-2.16)

0.0715287

c) If you select an individual stock from this population, the probability that it would have a return greater than -7 is

$=P(X>-7)$

$=1 - P(X\leq -7)$

1 -P( 2.6 10 -7- 2.6) X 10

$=1 - P(Z \leq -0.96)$

$=1 - \Phi(-0.96)$

d) Let $\bar{X}$ denotes the mean return for a random sample of four stocks from this population.

102 XNormal(2.6, 4
or $\bar{X}\sim Normal(2.6, 5^2)$

Now,

The probability that the sample would have a mean return less than 0 (a loss)

$=P(\bar{X}<0)$

$=P(\frac{\bar{X}-2.6}{5}<\frac{0-2.6}{5})$

$=P(Z<\frac{0-2.6}{5})$

$=\Phi(\frac{0-2.6}{5})$

$=\Phi(-0.52)$

e) The probability that the sample would have a mean return between -11 and -19

$=P(-19\leq \bar{X} \leq -11)$

$=P(\bar{X}\leq -11) - P(\bar{X}\leq -19)$

$=P(\frac{\bar{X}-2.6}{5}\leq \frac{-11-2.6}{5}) - P(\frac{\bar{X}-2.6}{5}\leq \frac{-19-2.6}{5})$

$= P(Z \leq -2.72)-P(Z \leq -4.32)$

D(-2.72 P(-4.32)

$\approx 0.0033$