The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1264 chips and a standard deviation of 116 chips. (a) Determine the 28th percentile for the number of chocolate chips in a bag. (b) Determine the number of chocolate chips in a bag that make up the middle 96% of bag.
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To solve both parts of this problem, we can use the standard score, also known as the z-score, which measures how many standard deviations an observation is from the mean. The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value of interest
- μ is the mean
- σ is the standard deviation
(a) To determine the 28th percentile for the number of chocolate chips in a bag, we need to find the value of x such that 28% of the data falls below it.
Step 1: Convert the percentile to a z-score.
Using a standard normal distribution table or calculator, find the z-score associated with the 28th percentile. The z-score corresponding to a percentile can also be found using the formula:
z = InvNorm(p)
Where p is the percentile.
Using p = 0.28, we find that z ≈ -0.591
Step 2: Calculate the value of x using the z-score formula.
Plug in the known values into the z-score formula:
-0.591 = (x - 1264) / 116
Now, solve for x:
-0.591 * 116 = x - 1264
-68.556 = x - 1264
x ≈ 1195.444
So, the 28th percentile for the number of chocolate chips in a bag is approximately 1195 chips.
(b) To determine the number of chocolate chips in a bag that make up the middle 96%, we need to find the range of x that represents the middle 96% of the data.
Step 1: Find the z-scores at the upper and lower percentiles.
To find the z-scores corresponding to the upper and lower percentiles, subtract and add half of the percentage that is outside the desired range from 1.
Lower z-score:
Z_lower = InvNorm((1 - 0.96) / 2) ≈ -1.75
Upper z-score:
Z_upper = InvNorm(1 - (1 - 0.96) / 2) ≈ 1.75
Step 2: Calculate the values of x using the z-score formula.
Plug in the known values into the z-score formula:
Z_lower = (x - 1264) / 116
Solve for x:
-1.75 = (x - 1264) / 116
-203 = x - 1264
x = 1061
Similarly, we can solve for the upper bound:
1.75 = (x - 1264) / 116
203 = x - 1264
x = 1467
So, the number of chocolate chips in a bag that make up the middle 96% is between 1061 and 1467 chips.
To solve this problem, we need to use the standard normal distribution table or a calculator that can give us the values for the z-scores.
(a) To determine the 28th percentile for the number of chocolate chips in a bag, we need to find the z-score associated with the 28th percentile using the standard normal distribution table.
The formula for the z-score is: z = (x - μ) / σ
Where:
- x is the value we want to convert to a z-score
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
In this case, we want to find the z-score for the 28th percentile, so we need to find the value of x that corresponds to the 28th percentile.
To find the z-score, we rearrange the formula as follows:
z = (x - μ) / σ
=> x = z * σ + μ
Substituting the values:
z = -0.611
σ = 116
μ = 1264
x = -0.611 * 116 + 1264
x ≈ 1194.88
Therefore, the 28th percentile for the number of chocolate chips in a bag is approximately 1195 chips.
(b) To determine the number of chocolate chips in a bag that make up the middle 96% of the bags, we need to find the z-scores that correspond to the cumulative probabilities at both ends.
The middle 96% corresponds to a cumulative probability of 0.96. This means that we want to find the z-scores associated with the 2nd and 98th percentiles.
Using the standard normal distribution table, we find that the z-score corresponding to the 2nd percentile is approximately -2.05 and the z-score corresponding to the 98th percentile is approximately 2.05.
Now we can use the formula x = z * σ + μ to find the values of x that correspond to these z-scores.
For the 2nd percentile:
x = -2.05 * 116 + 1264
x ≈ 1011.2
For the 98th percentile:
x = 2.05 * 116 + 1264
x ≈ 1515.4
Therefore, the number of chocolate chips in a bag that make up the middle 96% is approximately between 1011 and 1515 chips.