Mathematics, 23.03.2021 04:20 baller1216

The age of United States Presidents on the day of their first inauguration follows a Normal distribution with mean 56 and standard deviation 7.3. (a) (5 points) Compute the probability that a randomly selected President was less than 60 years old on the day of their first inauguration. (b) (5 points) Compute the 75th percentile for the age of United States Presidents on the day of inauguration. (c) (5 points) Compute the probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

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Answer from: sswd

a) 0.7088 = 70.88% probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

b) The 75th percentile for the age of United States Presidents on the day of inauguration is 61.

c) 0.8643 = 86.43% probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

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 pvalue, 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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The age of United States Presidents on the day of their first inauguration follows a Normal distribution with mean 56 and standard deviation 7.3.

This means that $\mu = 56, \sigma = 7.3$

(a) (5 points) Compute the probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

This is the pvalue of Z when X = 60. So

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

$Z = \frac{60 - 56}{7.3}$

Z = 0.55

Z = 0.55 has a pvalue of 0.7088

0.7088 = 70.88% probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

(b) (5 points) Compute the 75th percentile for the age of United States Presidents on the day of inauguration.

This is X when Z has a pvalue of 0.75. So X when Z = 0.675.

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

$0.675 = \frac{X - 56}{7.3}$

X - 56 = 0.675*7.3

X = 61

The 75th percentile for the age of United States Presidents on the day of inauguration is 61.

(c) (5 points) Compute the probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

Now, by the Central Limit Theorem, we have that $n = 4, s = \frac{7.3}{\sqrt{4}} = 3.65$

This is the pvalue of Z when X = 60. So

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{60 - 56}{3.65}$

Z = 1.1

Z = 1.1 has a pvalue of 0.8643

0.8643 = 86.43% probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

Answer from: Quest

96 39 cause 135 -96=39 im smart (i used a calculater)

Answer from: Quest

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