The weights of cereal boxes was found to be normally distributed. The mean is 16 oz and the standard deviation is 0.5 oz. From a sample of 250 boxes about how many should weigh between 15.5 oz and 17 oz?
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability between the two Z scores. Multiply by 250.
203 boxes
40
To solve this question, we can use the properties of the normal distribution.
First, let's define the parameters of the normal distribution:
Mean (μ) = 16 oz
Standard deviation (σ) = 0.5 oz
Now, we need to find the probability that a box weighs between 15.5 oz and 17 oz.
To find this probability, we can use the concept of the standardized normal distribution or Z-scores. The Z-score formula is given by:
Z = (X - μ) / σ
Where X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case, we want to find the probability for boxes weighing between 15.5 oz and 17 oz.
For 15.5 oz:
Z1 = (15.5 - 16) / 0.5 = -1
For 17 oz:
Z2 = (17 - 16) / 0.5 = 2
Now, we can use a Z-table or a calculator to find the probabilities corresponding to these Z-scores.
Looking up the Z-table, we find that the area under the normal distribution curve from Z = -1 to Z = 2 is approximately 0.8186.
Therefore, the probability that a box weighs between 15.5 oz and 17 oz is approximately 0.8186.
To find the number of boxes that would fall within this weight range, we multiply the probability by the total sample size (250):
Number of boxes ≈ 0.8186 * 250
Number of boxes ≈ 204.65
Rounded to the nearest whole number, approximately 205 boxes should weigh between 15.5 oz and 17 oz based on the given mean and standard deviation.