Suppose the average age of family members is 34 with a standard deviation of 4. if 100 members of the community decided to have a summer outing bonding and relaxation, find the probability that the average of these members is less than 35?

Respuesta :

Using the normal distribution, it is found that there is a 0.9938 = 99.38% probability that the average of these members is less than 35.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this problem, we have that the parameters are given as follows:

[tex]\mu = 34, \sigma = 4, n = 100, s = \frac{4}{\sqrt{100}} = 0.4[/tex].

The probability that the average of these members is less than 35 is the p-value of Z when X = 35, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{35 - 34}{0.4}[/tex]

Z = 2.5

Z = 2.5 has a p-value of 0.9938.

0.9938 = 99.38% probability that the average of these members is less than 35.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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