Respuesta :
Answer:
69.65 inches
Step-by-step explanation:
Mean = [tex]\mu = 65[/tex]
Standard deviation = [tex]\sigma = 2[/tex]
Now we are supposed to find if Rachael is at the 99th percentile in height for adult women, then her height, in inches
Formula : [tex]z =\frac{x-\mu}{\sigma}[/tex]
z at 99th percentile =2.326
Substitute the values in the formula
[tex]2.326=\frac{x-65}{2}[/tex]
[tex]2.326 \times 2=x-65[/tex]
[tex]4.652=x-65[/tex]
[tex]4.652+65=x[/tex]
[tex]69.65=x[/tex]
Hence her height is 69.65 inches .
Using the normal distribution, it is found that her height is of 69.7 inches.
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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]
- Each z-score has an associated p-value, which we find looking at the z-table, and represents the percentile of X.
In this problem:
- Mean of 65 inches, thus [tex]\mu = 65[/tex].
- Standard deviation of 2 inches, thus [tex]\sigma = 2[/tex].
- Rachael's height is at the 99th percentile, which means that her height is X when Z has a p-value of 0.99, so X when Z = 2.327.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.327 = \frac{X - 65}{2}[/tex]
[tex]X - 65 = 2(2.327)[/tex]
[tex]X = 69.7[/tex]
Her height is of 69.7 inches.
A similar problem is given at https://brainly.com/question/24887855
