Sugar packaged by a certain machine has a mean weight of 5 lb and a standard deviation of 0.04 lb. For what values of c can the manufacturer of the machinery claim that the sugar packaged by this machine has a weight between 5 − c and 5 + c lb with probability at least 84%. At least = c
= 0.08 lb
To find the value of c that satisfies the given condition, we can use the properties of a normal distribution.
First, let's understand that the mean weight of the sugar packaged by the machine is 5 lb with a standard deviation of 0.04 lb. This means that the weight of the sugar follows a normal distribution with a mean of 5 lb and a standard deviation of 0.04 lb.
Now, we need to find the value of c such that the weight of the sugar falls within the range of 5 - c and 5 + c lb with a probability of at least 84%.
To do this, we can use the properties of the standard normal distribution. Since we know the mean and standard deviation of the distribution, we can convert the given weights to standard deviations from the mean using the formula:
z = (x - μ) / σ,
where z is the number of standard deviations, x is the weight, μ is the mean, and σ is the standard deviation.
For the lower bound of 5 - c lb:
z1 = (5 - c - 5) / 0.04 = -c / 0.04 = -25c,
For the upper bound of 5 + c lb:
z2 = (5 + c - 5) / 0.04 = c / 0.04 = 25c.
We want the probability of the weight falling within this range to be at least 84%, which can be represented as:
P(-c/0.04 ≤ z ≤ c/0.04) ≥ 0.84.
Since the standard normal distribution is symmetrical, we can find the value of c by finding the z-value corresponding to the upper percentage point (1 - 0.16 / 2 = 0.92) using a standard normal distribution table or calculator.
The z-value for 0.92 is approximately 1.405.
Therefore, we have the inequality:
-25c ≤ 1.405 ≤ 25c.
Simplifying the inequality:
-25c ≤ 1.405,
c ≥ -1.405 / 25,
c ≥ -0.0562.
So, the manufacturer of the machinery can claim that the sugar packaged by this machine has a weight between 5 - c and 5 + c lb with a probability of at least 84% for c values greater than or equal to -0.0562.
To find the values of c, we need to determine the range within which the weight of the sugar packaged by the machine falls with a probability of at least 84%.
First, let's understand the probability distribution of the sugar weights. In this case, we have a normal distribution, as the mean weight and standard deviation are given. The mean weight is 5 lb and the standard deviation is 0.04 lb.
Next, we need to find the z-scores corresponding to the lower and upper limits of the weight range. The z-score represents the number of standard deviations a particular value is away from the mean in a normal distribution.
For the lower limit of 5 - c lb, we can calculate the z-score as follows:
z_lower = (weight - mean) / standard deviation
= (5 - (5 - c)) / 0.04
= c / 0.04
= 25c
Similarly, for the upper limit of 5 + c lb:
z_upper = (weight - mean) / standard deviation
= (5 - (5 + c)) / 0.04
= -c / 0.04
= -25c
To find the range within which the weight falls with a probability of at least 84%, we need to find the z-scores that correspond to the 8% tails on both sides of the distribution.
The critical z-scores can be found using a standard normal distribution table or a calculator. For an 84% probability, we need to find the z-scores that exclude 8% from each tail, resulting in a cumulative probability of 92%.
Thus, we need to find the z-score that corresponds to a cumulative probability of (1 - 0.08) / 2 = 0.46 (or 46% of the area under the curve).
Using a standard normal distribution table, the z-score corresponding to 46% is approximately 0.68.
Therefore, we equate the z-scores to ±0.68:
25c = 0.68
c = 0.68 / 25
c ≈ 0.0272
So, the manufacturer of the machinery can claim that the sugar packaged by this machine has a weight between 5 - 0.0272 lb and 5 + 0.0272 lb with a probability of at least 84%.