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torisob [31]
3 years ago
15

A corporation is considering a new issue of convertible bonds, mandagemnt belives that the offer terns will be founf attractiv b

y 20% of all its current stockholders, suppoer that the belif is correct, a random sample of 130 stockholderss is taken.
a. What is the standard error of the sample proportion who find this offer attractive?
b. What is the probability that the sample proportion is more than 0.15?
c. What is the probability that the sample proportion is between 0.18 and 0.22?
d. Suppose that a sample of 500 current stockholders had been taken. Without doing the calculations, state whether the probabilities in parts (b) and (c) would have been higher, lower, or the same as those found.

Mathematics
2 answers:
ella [17]3 years ago
4 0

Answer:

a. What is the standard error of the sample proportion who find this offer attractive? =  0.0351

b. What is the probability that the sample proportion is more than 0.15? = 0.9222

c. What is the probability that the sample proportion is between 0.18 and 0.22? =  0.4314

Step-by-step explanation:

see attachment for explanation

dezoksy [38]3 years ago
4 0

Answer:

Step-by-step explanation:

Given that,

p = 0.20

1 - p = 1 - 0.20 = 0.80

n = 130

\mu\hat p = p = 0.20

A) \sigma \hat p =  \sqrt[p( 1 - p ) / n] = \sqrt [(0.20 * 0.80 ) / 130] = 0.0351

B) P( \hat p > 0.15) = 1 - P( \hat p < 0.15)

= 1 - P(( \hat p - \mu \hat p ) / \sigma \hat p < (0.15 - 0.20) / 0.0351 )

= 1 - P(z < -1.42)

Using z table

= 1 - 0.0778

= 0.9222

C) P(0.18 < \hat p < 0.22)

= P[(0.18 - 0.20) / 0.0351 < ( \hat p - \mu \hat p ) / \sigma \hat p < (0.22 - 0.20) / 0.0351]

= P(-0.57 < z < 0.57)

= P(z < 0.57) - P(z < -0.57)

Using z table,    

= 0.7157 - 0.2843

= 0.4314

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The mailing list of an agency that markets scuba-diving trips to the Florida Keys contains 78% males and 22% females. The agency
choli [55]

Answer:

a) 0.98% probability that 17 of the 29 people are men.

b) 3.86% probability that the first woman is reached on the 8th call

Step-by-step explanation:

For each person chosen by the agency, there are only two possible outcomes. Either it is a man, or it is a woman. The probability of selecting a man or a women in each trial is independent from other trials. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

(a) What is the probability that 17 of the 29 people are men?

This is P(X = 17) when n = 29, p = 0.78. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 17) = C_{29,17}.(0.78)^{17}.(0.22)^{12} = 0.0098

0.98% probability that 17 of the 29 people are men.

(b) What is the probability that the first woman is reached on the 8th call?

On the first 7 trials, all men, each with a 78% probability.

On the 8th trial, a women, with a 22% probability. So

P = (0.78)^{7}*0.22 = 0.0386

3.86% probability that the first woman is reached on the 8th call

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Answer:

x ≥ 6 or x ≤ -2

Step-by-step explanation:

- I x - 2 I + 4 ≤ 0

- I x - 2 I ≤ -4

I x - 2 I ≥ 4

x - 2 ≥ 4  or  x - 2 ≤ -4

x ≥ 6  or x ≤ -2

5 0
3 years ago
I will mark you the brainiest for the correct answer, please be correct, Have a good day and take care, thanks. ( VIEW THE IMAGE
Gwar [14]
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