A normal distribution has a mean of 33 and a standard deviation of 4. Find the probability that a randomly selected x-value from the distribution is in the given interval. a. between 29 and 37 b. between 33 and 45 c. at least 29 d. at most 37

Respuesta :

Answer:

a. 0.6826 = 68.26% probability that a randomly selected x-value from the distribution is between 29 and 37.

b. 0.4987 = 49.87% probability that a randomly selected x-value from the distribution is between 33 and 45.

c. 0.8413 = 84.13% probability that a randomly selected x-value from the distribution is at least 29.

d. 0.8413 = 84.13% probability that a randomly selected x-value from the distribution is at most 37.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

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.

A normal distribution has a mean of 33 and a standard deviation of 4.

This means that [tex]\mu = 33, \sigma = 4[/tex]

a. between 29 and 37

This is the pvalue of Z when X = 37 subtracted by the pvalue of Z when X = 29. So

X = 37

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

[tex]Z = \frac{37 - 33}{4}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a pvalue of 0.8413

X = 29

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

[tex]Z = \frac{29 - 33}{4}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

0.6826 = 68.26% probability that a randomly selected x-value from the distribution is between 29 and 37.

b. between 33 and 45

pvalue of Z when X = 45 subtracted by the pvalue of Z when X = 33. So

X = 45

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

[tex]Z = \frac{45 - 33}{4}[/tex]

[tex]Z = 3[/tex]

[tex]Z = 3[/tex] has a pvalue of 0.9987

X = 33

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

[tex]Z = \frac{33 - 33}{4}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a pvalue of 0.5

0.9987 - 0.5 = 0.4987

0.4987 = 49.87% probability that a randomly selected x-value from the distribution is between 33 and 45.

c. at least 29

This is 1 subtracted by the pvalue of Z when X = 29. So

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

[tex]Z = \frac{29 - 33}{4}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a pvalue of 0.1587

1 - 0.1587 = 0.8413

0.8413 = 84.13% probability that a randomly selected x-value from the distribution is at least 29.

d. at most 37

This is the pvalue of Z when X = 37. So

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

[tex]Z = \frac{37 - 33}{4}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a pvalue of 0.8413

0.8413 = 84.13% probability that a randomly selected x-value from the distribution is at most 37.

Following are the solution to the given points:

Given:

[tex]\to Mean \ (\mu)= 33\\\\ \to \text{standard deviation}\ (\sigma) =4 \\\\[/tex]

To find:

find points=?

Solution:

For point a:

[tex]\to P (29 <x<37) = P(\frac{29-33}{4} \leq \frac{X-\mu}{\sigma} \leq \frac{37-33}{4} )[/tex]

                              [tex]= P(\frac{-4}{4} \leq \frac{X-\mu}{\sigma} \leq \frac{4}{4} )\\\\= P(-1 \leq z \leq 1 )\\\\= P(-1 < z <1 )\\\\= P(Z <+1) - P(Z <-1) \\\\=0.84134 - 0.15866\\\\= 0.68268 \\\\=68.268\%[/tex]

For point b:

[tex]\to P (33 <x<45) = P(\frac{33-33}{4} \leq \frac{X-\mu}{\sigma} \leq \frac{45-33}{4} )[/tex]

                 [tex]= P(\frac{0}{4} \leq \frac{X-\mu}{\sigma} \leq \frac{12}{4} )\\\\= P(\frac{0}{4} \leq \frac{X-\mu}{\sigma} \leq 3 )\\\\= P(-2< z< 0)\\\\= P(Z <-2) - P(Z <0) \\\\=0.9987-0.5 \\\\= 0.4987\\\\= 49.87\%[/tex]  

For point c:

When X = 29, subtract by 1 from the p-value of Z.

Using formula:

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

        [tex]=\frac{29-33}{4} \\\\ =\frac{-4}{4} \\\\= -1 \\\\=1 - 0.1587 \\\\= 0.8413 \\\\= 84.13\%[/tex]

For point d:

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

        [tex]= \frac{37-33}{4}\\\\ =1\\\\ =\text{1 has a p-value of}\ 0.8413\\\\=0.8413 \\\\= 84.13\%[/tex]

So, the answer is " 68.268%,49.87%, and 84.13%"

Learn more about the normal distribution:

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