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The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 7.8 cm. a. Find the probability that an individual distance is greater than 218.00 cm. b. Find the probability that the mean for 25 randomly selected distances is greater than 204.20 cm. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? a.

Respuesta :

Using the normal distribution, we have that:

a) There is a 0.0548 = 5.48% probability that an individual distance is greater than 218.00 cm.

b) There is a 0.7967 = 79.67% probability that the mean for 25 randomly selected distances is greater than 204.20 cm.

c) Since the underlying distribution is normal, the central limit theorem can be applied.

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], as long as the underlying distribution is normal or the sample size is greater than 30.

For this problem, the parameters are:

[tex]\mu = 205.5, \sigma = 7.8, n = 25, s = \frac{7.8}{\sqrt{25}} = 1.56[/tex]

Item a:

The probability is one subtracted by the p-value of Z when X = 218, hence:

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

[tex]Z = \frac{218 - 205.5}{7.8}[/tex]

Z = 1.6

Z = 1.6 has a p-value of 0.9452.

1 - 0.9452 = 0.0548.

There is a 0.0548 = 5.48% probability that an individual distance is greater than 218.00 cm.

Items b and c:

Since the underlying distribution is normal, the central limit theorem can be applied, hence the probability is one subtracted by the p-value of Z when X = 204.2.

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

By the Central Limit Theorem:

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

[tex]Z = \frac{204.2 - 205.5}{1.56}[/tex]

Z = -0.83

Z = -0.83 has a p-value of 0.2033.

1 - 0.2033 = 0.7967.

There is a 0.7967 = 79.67% probability that the mean for 25 randomly selected distances is greater than 204.20 cm.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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