The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 410 grams and a standard deviation of 10 grams. Find the weight that corresponds to each eventa. A.Highest 30 percent
b. Middle 70 percent to
c. Highest 90 percent
d.
Lowest 20 percent
a. Highest 30 percent: 441 grams
b. Middle 70 percent: 410 grams
c. Highest 90 percent: 441 grams
d. Lowest 20 percent: 381 grams
To find the weights corresponding to each of the given events, we can use the concept of z-scores in the standard normal distribution.
a. Highest 30 percent:
To find the weight corresponding to the highest 30 percent, we need to determine the z-score that corresponds to this percentile. We can use the z-table or a calculator to find the z-score.
Using the z-table, we look for the area of 0.30 in the body of the table. The closest value we can find is 0.2995, which corresponds to a z-score of approximately 0.52.
Next, we can use the formula z = (x - μ) / σ, where x is the weight we are trying to find, μ is the mean, and σ is the standard deviation.
Plugging in the values, we have:
0.52 = (x - 410) / 10
Rearranging the equation to solve for x:
10 * 0.52 = x - 410
5.2 + 410 = x
x = 415.2 grams
Therefore, the weight corresponding to the highest 30 percent is approximately 415.2 grams.
b. Middle 70 percent:
To find the weight corresponding to the middle 70 percent, we need to find the z-scores for the lower and upper percentiles of this range.
The lower percentile will be (100 - 70) / 2 = 15. The z-score corresponding to this area is approximately -0.88.
The upper percentile will also be 15. The corresponding z-score is approximately 0.88.
Using the formula z = (x - μ) / σ, we have two equations:
-0.88 = (x - 410) / 10 (lower z-score)
0.88 = (x - 410) / 10 (upper z-score)
Solving these equations, we get:
-0.88 * 10 = x - 410
-8.8 + 410 = x
x ≈ 401.2 grams (lower bound)
0.88 * 10 = x - 410
8.8 + 410 = x
x ≈ 418.8 grams (upper bound)
Therefore, the weight corresponding to the middle 70 percent is approximately between 401.2 grams and 418.8 grams.
c. Highest 90 percent:
To find the weight corresponding to the highest 90 percent, we use the same process as in part a.
Using the z-table, we find the closest value to 0.90, which is 0.8997. This corresponds to a z-score of approximately 1.28.
Using the formula z = (x - μ) / σ:
1.28 = (x - 410) / 10
Rearranging the equation to solve for x:
10 * 1.28 = x - 410
12.8 + 410 = x
x = 422.8 grams
Therefore, the weight corresponding to the highest 90 percent is approximately 422.8 grams.
d. Lowest 20 percent:
To find the weight corresponding to the lowest 20 percent, we use the same process as in part a.
The lowest 20 percent is equivalent to the highest 80 percent.
Using the z-table, we find the closest value to 0.80, which is 0.7967. This corresponds to a z-score of approximately 0.84.
Using the formula z = (x - μ) / σ:
0.84 = (x - 410) / 10
Rearranging the equation to solve for x:
10 * 0.84 = x - 410
8.4 + 410 = x
x = 418.4 grams
Therefore, the weight corresponding to the lowest 20 percent is approximately 418.4 grams.
To find the weight that corresponds to each event, we need to use the normal distribution and z-scores. A z-score measures how many standard deviations an observation or data point is from the mean.
a. Highest 30 percent:
To find the weight that corresponds to the highest 30 percent, we need to find the z-score that corresponds to the 30th percentile (or 0.30). We can use a standard normal distribution table or a z-score calculator to find this value.
Using a normal distribution table, we look for the closest value to 0.30, which is 0.3106. The corresponding z-score is approximately 0.525.
Next, we can use the formula for converting a z-score to an actual value:
Z = (X - μ ) / σ
Given:
μ = mean = 410 grams
σ = standard deviation = 10 grams
Z = 0.525
Plugging in the values:
0.525 = (X - 410) / 10
Solving for X, we get:
X = (0.525 * 10) + 410
X = 5.25 + 410
X = 415.25 grams
Therefore, the weight that corresponds to the highest 30 percent is approximately 415.25 grams.
b. Middle 70 percent:
To find the weight that corresponds to the middle 70 percent, we need to find the range between the 15th and 85th percentiles.
Using a standard normal distribution table (or z-score calculator), we find that the 15th percentile corresponds to a z-score of approximately -1.036 and the 85th percentile corresponds to a z-score of approximately 1.036.
Using the z-score formula:
Z = (X - μ ) / σ
For the 15th percentile:
-1.036 = (X - 410) / 10
For the 85th percentile:
1.036 = (X - 410) / 10
Solving for X in both equations, we get:
X = (-1.036 * 10) + 410 = 399.64 grams
X = (1.036 * 10) + 410 = 420.36 grams
Therefore, the weight that corresponds to the middle 70 percent is approximately 399.64 grams to 420.36 grams.
c. Highest 90 percent:
To find the weight that corresponds to the highest 90 percent, we need to find the z-score that corresponds to the 90th percentile (or 0.90).
Using a normal distribution table, we look for the closest value to 0.90, which is 0.8159. The corresponding z-score is approximately 1.282.
Using the z-score formula:
Z = (X - μ ) / σ
Given:
μ = mean = 410 grams
σ = standard deviation = 10 grams
Z = 1.282
Plugging in the values:
1.282 = (X - 410) / 10
Solving for X, we get:
X = (1.282 * 10) + 410
X = 12.82 + 410
X = 422.82 grams
Therefore, the weight that corresponds to the highest 90 percent is approximately 422.82 grams.
d. Lowest 20 percent:
To find the weight that corresponds to the lowest 20 percent, we need to find the z-score that corresponds to the 20th percentile (or 0.20).
Using a normal distribution table, we look for the closest value to 0.20, which is 0.2119. The corresponding z-score is approximately -0.842.
Using the z-score formula:
Z = (X - μ ) / σ
Given:
μ = mean = 410 grams
σ = standard deviation = 10 grams
Z = -0.842
Plugging in the values:
-0.842 = (X - 410) / 10
Solving for X, we get:
X = (-0.842 * 10) + 410
X = -8.42 + 410
X = 401.58 grams
Therefore, the weight that corresponds to the lowest 20 percent is approximately 401.58 grams.