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Some newly discovered vine has leaves that have a length that is normally distributed with mean 3 units, and standard deviation .2 (2/10). What percent of the leaves should be shorter than 3.5 units long?

Respuesta :

[tex]\text{ Answer: }P(Z<2.5)=0.99379=99.37\%[/tex]

Step-by-step explanation:

Since we have given that

Mean =3 units

Standard deviation =0.2

Using normal distribution, we need to find the percent of the leaves should be shorter than 3.5 units long,

[tex]P(Z<\frac{X-\mu}{\sigma})=P(Z<\frac{3.5-3}{0.2})=P(Z<2.5)[/tex]

Now, using the normal distribution table,

[tex]P(Z<2.5)=0.99379 \\=99.37\%[/tex]

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