Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

hip dysplasia is a condition of the hip socket that is very common in c…

Question

hip dysplasia is a condition 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 looks at a random sample of 50 labrador retrievers and records x = the number of labs with dysplasia in the sample.

(a) is the probability distribution of x approximately normal? justify your answer.

(b) calculate the mean and standard deviation of x.

(c) interpret the standard deviation of x.

(d) what is the probability that x is at least 5?

(e) what is the probability that x is at most 8?

(f) what is the probability that x is between 5 and 8, inclusive?

(g) would it be unusual to get a sample of 50 labrador retrievers in which 10 or more have hip dysplasia? justify your answer.

Explanation:

Step1: Identify Distribution Type

We have a binomial distribution here with \( n = 20 \) (sample size) and \( p = 0.116 \) (probability of hip dysplasia in Labrador retrievers). To check if it can be approximated by a normal distribution, we use the rules of thumb: \( np\geq5 \) and \( n(1 - p)\geq5 \).

Step2: Calculate \( np \) and \( n(1 - p) \)

First, calculate \( np \):
\( np = 20\times0.116 = 2.32 \)

Then, calculate \( n(1 - p) \):
\( n(1 - p)=20\times(1 - 0.116)=20\times0.884 = 17.68 \)

Step3: Check Normal Approximation Conditions

For a binomial distribution \( X\sim B(n,p) \), the normal approximation \( X\sim N(np,np(1 - p)) \) is appropriate when \( np\geq5 \) and \( n(1 - p)\geq5 \). Here, \( np = 2.32<5 \), so the normal approximation is not appropriate. The binomial distribution is not approximately normal in this case because the number of "successes" (dogs with dysplasia) is too small (since \( np < 5 \)).

Answer:

The probability distribution of \( X \) (number of Labs with dysplasia) is a binomial distribution (\( X\sim B(20, 0.116) \)) and is \(\boldsymbol{\text{not approximately normal}}\) because \( np = 2.32<5 \) (failing the normal approximation condition for binomial distributions).