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3 years ago

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Suppose that a stock is currently selling for $100. The change in the stock's price during the next year follows a normal random

variable with a mean of $10 and a standard deviation of $20. What is the probability (rounded to the nearest hundredth) that the stock will sell for $85 or less in a year's time?

1 answer:

Basile [38]3 years ago

6 0

Answer:

The probability that the stock will sell for $85 or less in a year's time is 0.10.

Step-by-step explanation:

Let <em>X</em> = stock's price during the next year.

The random variable <em>X</em> follows a normal distribution with mean, <em>μ</em> = $100 + $10 = $110 and standard deviation, <em>σ</em> = $20.

To compute the probability of a normally distributed random variable we first need to compute the <em>z</em>-score for the given value of the random variable.

The formula to compute the <em>z</em>-score is:

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

Compute the probability that the stock will sell for $85 or less in a year's time as follows:

Apply continuity correction:

P (X ≤ 85) = P (X < 85 - 0.50)

= P (X < 84.50)

$=P(\frac{X-mu}{\sigma}$

$=P(Z$

*Use a <em>z</em>-table for the probability.

Thus, the probability that the stock will sell for $85 or less in a year's time is 0.10.

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(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

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Step-by-step explanation:

Let <em>X</em> = number of senior professionals who thought that global warming is having a significant impact on the environment.

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(1)

Compute the value of $P(0.64$ as follows:

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Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 90%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of <em>z</em> for P (Z < z) = 0.95 is

<em>z</em> = 1.65.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73$

Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3)

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That is,

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Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

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