Question

The mean height of an adult giraffe is 18 feet. Suppose that the distribution is normally distributed with standard deviation 0.8 feet. Let X be the height of a randomly selected adult giraffe. Round all answers to two decimal places.What is the probability that a randomly selected giraffe will be shorter than 17 feet tall?What is the probability that a randomly selected giraffe will be between 16 and 19 feet tall? The 90th percentile for the height of giraffes is?

Answer 1

Answer:

1. the probability that giraffe will be shorter than 17 feet tall is equal to 0.1056

2. the probability that a randomly selected giraffe will be between 16 and 19 feet tall is equal to 0.8882

3. the the 90th percentile for the height of giraffes is 19.024

Step-by-step explanation:

To calculate the probability that a giraffe will be shorter than 17 feet tall, we need to standardize 17 feet as:

$z=\frac{x-m}{s}$

Where x is the height of the adult giraffe, m is the mean and s is the standard deviation, so 17 feet is equivalent to:

$z=\frac{17-18}{0.8}=-1.25$

Now, the probability that giraffe will be shorter than 17 feet tall is equal to P(z<-1.25). Then, using the standard normal distribution table, we get that:

$P(z<-1.25)=0.1056$

At the same way, 16 and 19 feet tall are equivalent to:

$z=\frac{16-18}{0.8}=-2.5\\z=\frac{19-18}{0.8}=1.25$

So, the probability that a randomly selected giraffe will be between 16 and 19 feet tall is equal to:

$P(-2.5<z<1.25)=P(z<1.25)-P(z<-2.5)\\P(-2.5<z<1.25)=0.8944-0.0062\\P(-2.5<z<1.25)=0.8882$

Finally, to find the the 90th percentile for the height of giraffes, we need to find the value z that satisfy:

$P(Z<z)=0.9$

Now, using the standard normal distribution table we get that z is equal to 1.28. Therefore, the height x of the giraffes that is equivalent to 1.28 is:

$z=\frac{x-m}{s} \\1.28=\frac{x-18}{0.8} \\x=(1.28*0.8)+18\\x=19.024$

it means that the the 90th percentile for the height of giraffes is 19.024