Using the normal distribution, it is found that 16.43% of adults in the US qualify for stage 1.
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Researching the problem on the internet, the mean and the standard deviation are given, respectively, by:
[tex]\mu = 122, \sigma = 22[/tex].
The proportion of adults in the US qualify for stage 1 is the p-value of Z when X = 160 subtracted by the p-value of Z when X = 140, hence:
X = 160:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{160 - 122}{22}[/tex]
Z = 1.73
Z = 1.73 has a p-value of 0.9582.
X = 140:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{140 - 122}{22}[/tex]
Z = 0.82
Z = 0.82 has a p-value of 0.7939.
0.9582 - 0.7939 = 0.1643.
0.1643 = 16.43% of adults in the US qualify for stage 1.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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