Respuesta :
Part A:
The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:
[tex]P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)[/tex]
Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:
[tex]P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057} [/tex]
Part B:
The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:
[tex]P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)[/tex]
Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:
[tex]P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}[/tex]
The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:
[tex]P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)[/tex]
Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:
[tex]P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057} [/tex]
Part B:
The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:
[tex]P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)[/tex]
Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:
[tex]P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}[/tex]
Using the normal distribution and the central limit theorem, it is found that there is a:
- a) 0.7054 = 70.54% probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months.
- b) 0.8869 = 88.69% probability that five randomly selected years will have an average precipitation greater than 18 inches for the first 7 months.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, 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:
- The mean is of 19.32 inches, hence [tex]\mu = 19.32[/tex].
- The standard deviation is of 2.44 inches, hence [tex]\sigma = 2.44[/tex].
Item a:
The probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is 1 subtracted by the p-value of Z when X = 18, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{18 - 19.32}{2.44}[/tex]
[tex]Z = -0.54[/tex]
[tex]Z = -0.54[/tex] has a p-value of 0.2946.
1 - 0.2946 = 0.7054.
0.7054 = 70.54% probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months.
Item b:
Now, we want the probability that five randomly selected years will have an average precipitation greater than 18 inches for the first 7 months, hence:
[tex]n = 5, s = \frac{2.44}{\sqrt{5}}[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{18 - 19.32}{\frac{2.44}{\sqrt{5}}}[/tex]
[tex]Z = -1.21[/tex]
[tex]Z = -1.21[/tex] has a p-value of 0.1131.
1 - 0.1131 = 0.8869.
0.8869 = 88.69% probability that five randomly selected years will have an average precipitation greater than 18 inches for the first 7 months.
To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213
