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.
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.
2 answers:
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
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Answer:
a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.
b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.
c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The mean of a population is 74 and the standard deviation is 15.
This means that
Question a:
Sample of 36 means that
This probability is 1 subtracted by the pvalue of Z when X = 78. So
By the Central Limit Theorem
has a pvalue of 0.9452
1 - 0.9452 = 0.0548
0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.
Question b:
Sample of 150 means that
This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So
X = 77
has a pvalue of 0.9929
X = 71
has a pvalue of 0.0071
0.9929 - 0.0071 = 0.9858
0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.
c. A random sample of size 219 yielding a sample mean of less than 74.2
Sample size of 219 means that
This probability is the pvalue of Z when X = 74.2. So
has a pvalue of 0.5793
0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2