Question

dysplasia is a malformation of the hip socket that is very common in certain dog breeds and causes arthritis as a dog gets older. according to the orthopedic foundation for animals, 11.6% of all labrador retrievers have hip dysplasia. a veterinarian tests a random sample of 50 labrador retrievers and records ( x = ) the number of labs with dysplasia in the sample.
is the probability distribution of ( x ) approximately normal? justify your answer.
( np = ) (\boxed{}) and ( n(1 - p) = ) (\boxed{})
therefore, the probability distribution of ( x ) (\boxed{}) approximately normal.

Explanation:

Step1: Identify values of n and p

We know that \( n = 50 \) (sample size) and \( p=0.116 \) (probability of a Labrador having dysplasia).

Step2: Calculate \( np \)

To find \( np \), we multiply \( n \) and \( p \):
\( np = 50\times0.116 = 5.8 \)

Step3: Calculate \( n(1 - p) \)

First, find \( 1 - p = 1 - 0.116 = 0.884 \). Then multiply by \( n \):
\( n(1 - p)=50\times0.884 = 44.2 \)

Step4: Check normal approximation conditions

For a binomial distribution \( X \sim \text{Binomial}(n,p) \), the normal approximation is appropriate if \( np \geq 5 \) and \( n(1 - p)\geq 5 \). Here, \( np = 5.8\geq5 \) and \( n(1 - p)=44.2\geq5 \), so the conditions are met.

Answer:

\( np = \boldsymbol{5.8} \) and \( n(1 - p)=\boldsymbol{44.2} \). Since \( np \geq 5 \) and \( n(1 - p)\geq 5 \), the probability distribution of \( X \) \(\boldsymbol{\text{is}}\) approximately normal.