Question

A certain mutual fund invests in both U.S. and foreign markets. Let x be a random variable that represents the monthly percentage return for the fund. Assume x has mean u = 1.8% and standard deviation o = 0.6%. (a) The fund has over 500 stocks that combine together to give the overall monthly percentage return x. We can consider the monthly return of the stocks in the fund to be a sample from the population of monthly returns of all world stocks. Then we see that the overall monthly return x for the fund is itself an average return computed using all 500 stocks in the fund. Why would this indicate that x has an approximately normal distribution? Explain. Hint: See the discussion after Theorem 6.2. The random variable x is a mean of a sample size n = 500. By the central limit theorem , the X v distribution is approximately normal. (b) After 6 months, what is the probability that the average monthly percentage return will be between 1% and 2%? Hint: See Theorem 6.1, and assume that x has a normal distribution as based on part (a). (Round your answer to four decimal places.) (c) After 2 years, what is the probability that will be between 1% and 2%? (Round your answer to four decimal places.)

Answer 1

Given :

A doe that weighs less than 60 kgs is considered malnourished.

Solution :

a)

$\mu = 69, \sigma =7.4$

The probability that a single doe captured at random in december is undernourished can be given as

T- P(<60) = P 60 – 69 7.4

PC <60) = P(Z < -1.2162)

b)

If the park has about 2600 does, the number of does that are expected to be undernourished in December are

$np = 2600*0.1131 = 294.06\approx 294$

c)

$\mu = 69, \sigma =7.4, n =45$

The probability that the average weight for a random sample of 45 does is less than 66 kg is

$P(x < 66) = P\left ( \frac{\bar{x} - \mu }{\frac{\sigma}{\sqrt{n}}} < \frac{66-69}{\frac{7.4}{\sqrt{45}}} \right )$

$P(x < 66) = P\left ( z < -2.72 \right )$

d)

$\mu = 69, \sigma =7.4, n =45$

The probability that x̅ < 70.3 kg for 45 does is

$P(x < 70.3) = P\left ( \frac{\bar{x} - \mu }{\frac{\sigma}{\sqrt{n}}} < \frac{70.3-69}{\frac{7.4}{\sqrt{45}}} \right )$

$P(x < 70.3) = P(z < 1.1786) \approx P(z < 1.18)$

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