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Assume the weights of painkiller pills are normally distributed with a mean of 550 mg and a standard deviation of 75 mg. If 81 pills are randomly selected, find the probability that they have a mean weight between 535 mg and 560 mg.

User Kina
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Final answer:

The probability that the mean weight of 81 randomly selected painkiller pills is between 535 mg and 560 mg is approximately 84.90%. This is calculated using the standard normal distribution and the z-score formula.

Step-by-step explanation:

To find the probability that the mean weight of 81 randomly selected painkiller pills falls between 535 mg and 560 mg, we must first consider the sampling distribution of the sample mean. Since the population distribution of weights is normally distributed with a mean (μ) of 550 mg and a standard deviation (σ) of 75 mg, the sampling distribution of the sample mean for n = 81 pills can also be assumed to be normal due to the Central Limit Theorem. Additionally, the standard error of the mean (SEM) is σ/√n, which in this case is 75 mg/√81 = 75 mg/9 = 8.33 mg (rounded to two decimal places).

Next, we can use the z-score formula to convert the values of 535 mg and 560 mg into z-scores. The z-score formula is Z = (X - μ) / SEM, where X is the value of interest:

  • Z for 535 mg = (535 - 550) / 8.33 ≈ -1.80
  • Z for 560 mg = (560 - 550) / 8.33 ≈ +1.20

Now, we look up the z-scores in the standard normal distribution table (or use a calculator or software with this functionality) to get the probabilities:

  • Probability of z ≤ -1.80 is approximately 0.0359
  • Probability of z ≤ +1.20 is approximately 0.8849

The probability that the sample mean falls between these two z-scores is:

Probability (535 mg < X < 560 mg) = P(z ≤ +1.20) - P(z ≤ -1.80) = 0.8849 - 0.0359 = 0.8490

The probability that the mean weight of the pills is between 535 mg and 560 mg is approximately 84.90%.

User Frilla
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