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Answer:
Jennifer's height is 63.7 inches.
Step-by-step explanation:
Let X = heights of adult women in the United States.
The random variable X is normally distributed with mean, μ = 65 inches and standard deviation σ = 2.4 inches.
To compute the probability of a normal random variable we first need to convert the raw score to a standardized score or z-score.
The standardized score of a raw score X is:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
These standardized scores follows a normal distribution with mean 0 and variance 1.
It is provided that Jennifer is taller than 70% of the population of U.S. women.
Let Jennifer's height be denoted by x.
Then according to the information given:
P (X > x) = 0.70
1 - P (X < x) = 0.70
P (X < x) = 0.30
⇒ P (Z < z) = 0.30
The z-score related to the probability above is:
z = -0.5244
*Use a z-table.
Compute the value of x as follows:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
[tex]-0.5244=\frac{x-65}{2.4}[/tex]
[tex]x=65-(0.5244\times 2.4)[/tex]
[tex]=63.7414\\\approx63.7[/tex]
Thus, Jennifer's height is 63.7 inches.
The required answer is [tex]63.7[/tex] inches.
Z-Score:
z score is simply defined as the number of standard deviation from the mean.
Given that,
[tex]\mu=65\\\sigma=2.4[/tex]
Taller than [tex]70\% = 0.70[/tex]
[tex]P(Z > z ) = 0.70\\1- P(Z < z) =0.70\\P(Z < z) = 1-0.70\\ = 0.30\\z = -0.524[/tex]
Using the z-score formula,
[tex]x=z\times \sigma+\mu\\x=-0.524\times 2.4+65\\x=63.7[/tex]
Learn more about the topic Z-Score:
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