Using the normal distribution, it is found that 95.65% of full term newborn female infants with a head circumference between 31 cm and 36 cm.
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]
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 33.8, \sigma = 1.2[/tex]
The proportion of full term newborn female infants with a head circumference between 31 cm and 36 cm is the p-value of Z when X = 36 subtracted by the p-value of Z when X = 31, hence:
X = 36:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{36 - 33.8}{1.2}[/tex]
Z = 1.83
Z = 1.83 has a p-value of 0.9664.
X = 31:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{31 - 33.8}{1.2}[/tex]
Z = -2.33
Z = -2.33 has a p-value of 0.0099.
0.9664 - 0.0099 = 0.9565.
0.9565 = 95.65% of full term newborn female infants with a head circumference between 31 cm and 36 cm.
More can be learned about the normal distribution at https://brainly.com/question/24537145
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