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Suppose the ages of multiple birth (3 or more babies) are normally distributed with a mean age of 31.7 years and a standard deviation of 5.2 years. What percent of these mothers are between the ages 30-35

Respuesta :

Answer:

The percent of these mothers are between the ages 30-35 is 36.53%

Step-by-step explanation:

we are given

mean of age =31.7 years

[tex]\mu=31.7[/tex]

standard deviation of 5.2 years

[tex]\sigma=5.2[/tex]

For age=30 years:

x=30

we can find z-score

[tex]z=\frac{x-\mu}{\sigma}[/tex]

we can plug values

[tex]z=\frac{30-31.7}{5.2}[/tex]

[tex]z=-0.32692[/tex]

For age=35 years:

x=35

we can find z-score

[tex]z=\frac{x-\mu}{\sigma}[/tex]

we can plug values

[tex]z=\frac{35-31.7}{5.2}[/tex]

[tex]z=0.63462[/tex]

now, we can use normal distribution table

[tex]P(-0.32692<z<0.63462)=0.3653[/tex]

now, we can find percentage

[tex]=0.3653\times 100[/tex]

=36.53%


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