Question

In: Statistics and Probability

The mean height of an adult giraffe is 17 feet. Suppose that the distribution is normally...

The mean height of an adult giraffe is 17 feet. Suppose that the distribution is normally distributed with standard deviation 0.9 feet. Let X be the height of a randomly selected adult giraffe. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(___,______)

b. What is the median giraffe height? _____ft.

c. What is the Z-score for a giraffe that is 19 foot tall?

d. What is the probability that a randomly selected giraffe will be shorter than 17.8 feet tall?

e. What is the probability that a randomly selected giraffe will be between 17.4 and 18.2 feet tall?

f. The 70th percentile for the height of giraffes is ________ft.

Solutions

Expert Solution

(a)

X N (17, 0.81)

X is Normal Distribution with mean = = 17 and Variance = = 0.81

(b)

Median of giraffe height = 17 ft.

(c)

X = 19

Z = (19 - 17)/0.9 = 2.2222

(d)

To find P(X<17.8):

Z = (17.8 - 17)/0.9 = 0.8889

Table of Area Under Standard Normal Curve gives area = 0.3133

So,

P(X<17.8) = 0.5 + 0.3133 = 0.8133

So,

Answer is:

0.8133

(e)

To find P(17.4<X<18.2):

Case 1: For X from mid value to 17.4:

Z = (17.4 - 17)/0.9 = 0.4444

Table of Area Under Standard Normal Curve gives area = 0.1700

Case 2: For X from mid value to 18.2:

Z = (18.2 - 17)/0.9 = 1.3333

Table of Area Under Standard Normal Curve gives area = 0.4082

So,

P(17.4<X<18.2) = 0.4082 - 0.1700 = 0.2382

So,

Answer is:

0.2382

(f)

70th percentile is given by area = 0.70 - 0.50 = 0.20 from mid value to Z on RHS.

Table gives Z = 0.525

So,

Z = 0.525 = (X - 17)/0.9

So,

X = 17 + (0.525 X 0.9)

= 17.4725

So,

Answer is:

17.4725


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