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Sagot :
Using the normal distribution, it is found that there is a 0.8224 = 82.24% probability that a randomly selected sunglasses will have a price between 63 dollars and 90 dollars.
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.
In this problem:
- Mean of 76, thus, [tex]\mu = 76[/tex].
- Standard deviation of 10, thus, [tex]\sigma = 10[/tex].
The probability of a price between 63 dollars and 90 dollars is the p-value of Z when X = 90 subtracted by the p-value of Z when X = 63, hence:
X = 90:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{90 - 76}{10}[/tex]
[tex]Z = 1.4[/tex]
[tex]Z = 1.4[/tex] has a p-value of 0.9192.
X = 63:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{63 - 76}{10}[/tex]
[tex]Z = -1.3[/tex]
[tex]Z = -1.3[/tex] has a p-value of 0.0968.
0.9192 - 0.0968 = 0.8224.
0.8224 = 82.24% probability that a randomly selected sunglasses will have a price between 63 dollars and 90 dollars.
A similar problem is given at https://brainly.com/question/24663213
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