what is the likelihood a pack of gummy bears will contain between 37 and 45 candies?

Question

the number of pieces in a pack of gummy bears is normally distributed with a mean of 35 and a standard deviation of 4. if a package is chosen at random what is the likelihood it will contain between 37 and 45 candies?

Answer 1

First, you need to determine the z-scores of a package containing 37 and 45 candies. The z-scores for each situation will explain how many standard deviations above/below the mean of 35. Notice that both values are above 35, this means that both z-scores will be positive, since both values are above the mean.

The formula for calculating a z-score is: z-score = (observed value - mean)/standard deviation.

The z-score for 37 is (37 - 35)/4 = 0.5.

The z-score for 45 is (45 - 35)/4 = 2.5.

Now we need to find the probability that a value will fall between 0.5 standard deviations above a mean and 2.5 standard deviations above a mean. To do this we must refer to "Table A" in a statistics textbook, regarding z-scores. The z-score of 0.5 yields a value of 0.6915, and the z-score of 2.5 yields a value of 0.9938. It is important to remember that these decimal values given to us in Table A are probabilities of events occurring at or below a given z-score. This means that everything under the normal curve to the left of z = 0.5 has ben counted twice (since it is also part of the space to the left of z - 2.5). Therefore, we must subtract the value for z = 0.5 from the value for z = 2.5, so that we are only counting the area in between the two values.

Thus, 0.9938 - 0.6915 = 0.3023. In conclusion, there is a 30.23% chance of a package of gummy bears having between 37 and 45 candies if the average is 35 and the standard deviation is 4.