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Umnica [9.8K]
2 years ago
12

4 letters are typed, with repetition allowed. What is the probability that all 4 will be vowels? Write your answer as apercent.

Round to the nearest hundredth of a percent as needed.The probability is%.Check Answer
Mathematics
1 answer:
zubka84 [21]2 years ago
7 0

The total number of alphabets are 26.

The total number of vowels are 5.

Th repetition is allowed.

The possible outocme for all to be vowels are,

PO=5^4PO=625

The total outcome is,

TO=26^4

Therefore, the probability that all 4 will be vowels ,

P=\frac{625}{456976}P=0.00137

The probability is 0.137 %.

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Answer:

(a) (2.573, 3.167) is the 99% confidence interval for \mu the true mean calcium level of the population of people who experienced an unexplained episode of vitamin D intoxication.

(b) There is a 99% probability that the average calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.

Step-by-step explanation:

If we have a random sample of size n from a normal distribution with mean \mu and standard deviation \sigma, then we know that \bar{X} is normally distributed with mean \mu and standard deviation \sigma/n. Therefore we can use (\bar{X}-\mu)/\frac{\sigma}{\sqrt{n}} as a pivotal quantity.

We have a sample size of n = 12. The sample mean is \bar{x} = 2.87 mmol/l and the standard deviation is \sigma = 0.40. The confidence interval is given by \bar{x}\pm z_{\alpha/2}(\frac{\sigma}{\sqrt{n}}) where z_{\alpha/2} is the \alpha/2th quantile of the standard normal distribution. As we want the 99% confidence interval, we have that \alpha = 0.01 and the confidence interval is 2.87\pm z_{0.005}(\frac{0.40}{\sqrt{12}}) where z_{0.005} is the 0.5th quantile of the standard normal distribution, i.e., z_{0.005} = -2.5758. Then, we have 2.87\pm (-2.5758)(\frac{0.40}{\sqrt{12}}) and the 99% confidence interval is given by (2.573, 3.167)

(b) We found a 99% confidence interval for the true mean calcium level of the population of people who experienced an unexplained episode of vitamin D intoxication. Therefore, there is a 99% probability that the average calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.

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Step-by-step explanation:

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