Question

We have heard rumors of vicious honey badger attacks in the forests of northern Michigan. In an effort to clear the good name of honey badgers everywhere, we do some research and discover that nation wide there is about 0.5% of being attacked by a honey badger. We take a sample of 1100 honey badgers. Can we use an approximately normal distribution? Find the probability of having at most 6 honey badgers involved in an attack on a person?

Suppose we followed our honey badger sample for 4 years and determined that there was 4 instances of attackers on humans from our sample. Find p. Find the standard error for our sample. If it is quite likely for the true proportion of honey badger attacks to be within 2 standard errors of our sample proportion, find an interval that is likely to contain the true population proportion of honey badger attacks.

Answer 1

Solution:

We are given that: In an effort to clear the good name of honey badgers everywhere, we do some research and discover that nation wide there is about 0.5% of being attacked by a honey badger.

That is : p = probability of being attacked by a honey badger = 0.5% = 0.005

Sample size = n = 1100

Part a) Can we use an approximately normal distribution?

Here X = number of honey badgers involved in an attack on a person follows Binomial distribution with parameters n = 1100 and p = 0.005

Since sample size is large, so we can use an approximately normal distribution if following conditions satisfied.

   and

and

Since both the conditions are satisfied, we can use an approximately normal distribution.

Part b) Find the probability of having at most 6 honey badgers involved in an attack on a person?

That is we have to find:

Since we are using an approximately normal distribution, we need to use continuity correction.

That is we add or subtract 0.5 from x value.

Here inequality is Less than or equal to , so in order to include 6 in the range of x values, we must add 0.5, thus we get:

Now find z score:

where

Mean:

Standard Deviation:

Thus we get:

Thus we get:

Look in z table for z = 0.4 and 0.03 and find corresponding area.

P( Z < 0.43) = 0.6664

Thus

Thus the probability of having at most 6 honey badgers involved in an attack on a person is 0.6664