The Acme Candy Company claims that 60% of the jawbreakers it produces weigh more than .4ounces. Suppose that 800 jawbreakers are selected at random from the production lines. Would it be
unusual for this sample of 800 to contain 494 jawbreakers that weigh more than .4 ounces?
yes
any math to support this response?
To determine if it would be unusual for the sample to contain 494 jawbreakers that weigh more than 0.4 ounces, we need to calculate the probability.
We can use the normal approximation to the binomial distribution since the sample size is large (n = 800) and the probability of success (p = 0.6) is not too close to 0 or 1.
The mean of the binomial distribution is μ = np, and the standard deviation is σ = sqrt(np(1-p)).
Calculating the mean and standard deviation:
Mean (μ) = 800 * 0.6 = 480
Standard Deviation (σ) = sqrt(800 * 0.6 * (1-0.6)) = sqrt(192)
Now we can convert the given value (494 jawbreakers) into a z-score using the formula:
z = (x - μ) / σ
Where x is the given value.
Calculating the z-score:
z = (494 - 480) / sqrt(192) ≈ 1.37
To determine if this outcome is unusual, we can compare the z-score to the standard normal distribution, noting that a z-score above 3 or below -3 is generally considered unusual.
Since the calculated z-score of 1.37 is less than 3, it is not considered unusual for the sample of 800 jawbreakers to contain 494 jawbreakers that weigh more than 0.4 ounces.
To determine if it would be unusual for the sample of 800 jawbreakers to contain 494 that weigh more than 0.4 ounces, we need to calculate the probability of observing this outcome.
We can use the binomial probability formula to calculate the probability of obtaining a specific number of successes in a given number of trials. In this case, the probability can be calculated using the following formula:
P(X = k) = nCk * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of observing exactly k successes,
n is the number of trials (sample size),
k is the number of successes in the sample,
p is the probability of success in a single trial (probability that a randomly selected jawbreaker weighs more than 0.4 ounces), and
nCk is the binomial coefficient (number of ways to choose k successes from n trials).
In this scenario, the probability of a jawbreaker weighing more than 0.4 ounces is 60% or 0.6, and the sample size is 800. So, we can calculate the probability of observing exactly 494 jawbreakers weighing more than 0.4 ounces using the binomial probability formula:
P(X = 494) = 800C494 * 0.6^494 * (1 - 0.6)^(800 - 494)
To calculate this, we need to evaluate the expression for the binomial coefficient, raise 0.6 to the power of 494, raise 0.4 to the power of (800 - 494), and multiply these values by each other. However, calculating the binomial coefficient and powers manually can be time-consuming.
Alternatively, we can use statistical software or an online tool to calculate the binomial probability. For instance, using a binomial probability calculator, we can input the values n = 800, k = 494, and p = 0.6 to obtain the probability directly.
If the resulting probability is relatively small (usually less than 0.05), we consider it unusual or statistically significant. If the probability is larger than 0.05, we would consider it not unusual or statistically insignificant.
Therefore, by calculating the probability, or using a calculator, if the probability is less than 0.05, it would be considered unusual, and if it is greater than 0.05, it would not be considered unusual.