The ratio of brown M&M’s in a bag is stated by the candy company to be
56%. If a random sample of 500 bags is selected, determine the probability that
the sample proportion of random bags of M&M’s selected has a percentage of
brown between 51% and 59%
p=.56
standard deviation=sqrt(.56(1-.56)/500)= .0222
Since you have the mean and sd
plug in your range into
http://davidmlane.com/hyperstat/z_table.html
(one of the best sites for replacing all those nasty tables in the back of your text)
To determine the probability that the sample proportion of random bags of M&M's selected has a percentage of brown between 51% and 59%, we need to use the standard deviation and the mean.
The mean of the sample proportion will be the same as the population proportion, which is p = 0.56.
The standard deviation of the sample proportion is calculated using the formula sqrt(p(1-p)/n), where p is the population proportion (0.56) and n is the sample size (500). Plugging in the values, we get sqrt(0.56(1-0.56)/500) = 0.0222.
Now, to find the probability of the sample proportion falling between 51% and 59%, we need to convert these percentages to proportions.
51% as a proportion is 0.51, and 59% as a proportion is 0.59.
To find the probability, we will use the standard normal distribution since the sample size is large (n > 30) and the data is binomially distributed.
Step 1: Find the z-scores for the given values.
For 51%, the z-score is (0.51 - p) / standard deviation = (0.51 - 0.56) / 0.0222 = -2.25
For 59%, the z-score is (0.59 - p) / standard deviation = (0.59 - 0.56) / 0.0222 = 1.35
Step 2: Use a standard normal distribution table or a calculator to find the area under the curve between these two z-scores.
Let's assume an approximate value using a standard normal distribution table:
Area to the left of z = -2.25 is approximately 0.0122.
Area to the left of z = 1.35 is approximately 0.9115.
Step 3: Find the probability by subtracting the area to the left of z = -2.25 from the area to the left of z = 1.35.
Probability = 0.9115 - 0.0122 = 0.8993
So, the probability that the sample proportion of random bags of M&M's selected has a percentage of brown between 51% and 59% is approximately 0.8993 or 89.93%.