The actual weights of bag of pet food are normally distributed.The mean of the weights is 50.0 lb,with a standard deviation of 0.2 lb.
C)In a group of 50 bags,how many would yu expect to be within 1.5 standard deviations of the mean?
25
To find out how many bags you would expect to be within 1.5 standard deviations of the mean, you need to calculate the z-score for the value that represents 1.5 standard deviations above and below the mean.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value you want to find the z-score for
- μ is the mean
- σ is the standard deviation
For 1.5 standard deviations above the mean:
x1 = μ + (1.5 * σ)
And for 1.5 standard deviations below the mean:
x2 = μ - (1.5 * σ)
Substituting the values given in the question:
x1 = 50.0 + (1.5 * 0.2)
x1 = 50.0 + 0.3
x1 = 50.3
x2 = 50.0 - (1.5 * 0.2)
x2 = 50.0 - 0.3
x2 = 49.7
Now that we have the limits, we can calculate the z-scores:
z1 = (x1 - μ) / σ
z2 = (x2 - μ) / σ
Substituting the values:
z1 = (50.3 - 50.0) / 0.2
z1 = 0.3 / 0.2
z1 = 1.5
z2 = (49.7 - 50.0) / 0.2
z2 = -0.3 / 0.2
z2 = -1.5
Next, we need to find the area under the standard normal distribution curve between these two z-scores. This can be done using a standard normal distribution table or a calculator.
For z = 1.5, the area under the curve is approximately 0.9332.
For z = -1.5, the area under the curve is also approximately 0.9332.
To find the proportion of bags within 1.5 standard deviations of the mean, we subtract the area outside this range (1 - 2 * 0.9332):
P(-1.5 < z < 1.5) ≈ 1 - 2 * 0.9332
P(-1.5 < z < 1.5) ≈ 1 - 1.8664
P(-1.5 < z < 1.5) ≈ 0.1336
Now, we multiply this proportion by the total number of bags (50) to find the expected number of bags within 1.5 standard deviations of the mean:
Expected number = Proportion * Total number
Expected number = 0.1336 * 50
Expected number ≈ 6.68
Therefore, you would expect approximately 6.68 bags to be within 1.5 standard deviations of the mean in a group of 50 bags.