A scientist studying water quality measures the lead level in parts per billion (ppb) at each of 49 randomly chosen locations along a water line. Suppose that the lead levels across all the locations on this line are strongly skewed to the right with a mean of $17 \mathrm{ppb}$ and a standard deviation of $14 \mathrm{ppb}$. Assume that the measurements in the sample are independent.

What is the probability that the mean lead level from the sample of 49 measurements $\bar{x}$ is less than $15 \mathrm{ppb}$ ?

Choose 1 answer:
(A) $P(\bar{x}

Q: A scientist studying water quality measures the lead level in parts per billion (ppb) at each of 49 randomly chosen locations along a water line. Suppose that the lead levels across all the locations on this line are strongly skewed to the right with a mean of $17 \mathrm{ppb}$ and a standard deviation of $14 \mathrm{ppb}$. Assume that the measurements in the sample are independent. What is the probability that the mean lead level from the sample of 49 measurements $\bar{x}$ is less than $15 \mathrm{ppb}$ ? Choose 1 answer: (A) $P(\bar{x}<15) \approx 0.02$ (B) $P(\bar{x}<15) \approx 0.16$ (C) $P(\bar{x}<15) \approx 0.30$ (D) $P(\bar{x}<15) \approx 0.44$ (E) We cannot calculate this probability because the sampling distribution is not normal.

∻Solution by Steps∻ ‖ step 1 ⋮ To find the probability that the sample mean $\bar{x}$ is less than 15 ppb, we use the standard normal distribution (Z-distribution). ‖ ‖ step 2 ⋮ We calculate the Z-score for $\bar{x} = 15$ using the formula $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\mu = 17$ ppb is the population mean, $\sigma = 14$ ppb is the population standard deviation, and $n = 49$ is the sample size. ‖ ‖ step 3 ⋮ Substituting the values into the Z-score formula gives $Z = \frac{15 - 17}{14/\sqrt{49}} = \frac{-2}{2} = -1$. ‖ ‖ step 4 ⋮ We look up the Z-score of -1 in the standard normal distribution table or use a calculator to find the cumulative probability. ‖ ‖ step 5 ⋮ The cumulative probability for Z = -1 is approximately 0.16. This is the probability that the sample mean $\bar{x}$ is less than 15 ppb. ‖ ***B*** ∻Key Concept∻ ⚹Central Limit Theorem⚹ ∻Explanation∻ ⚹The Central Limit Theorem states that the sampling distribution of the sample mean will be approximately normally distributed, regardless of the population distribution, provided the sample size is sufficiently large. In this case, with a sample size of 49, we can assume the sampling distribution of the mean is approximately normal even though the population distribution is skewed.⚹◊What is the formula to calculate the probability that the mean lead level from the sample of 49 measurements is less than 15 ppb?⍭ Generate me a similar question◊

Question

A scientist studying water quality measures the lead level in parts per billion (ppb) at each of 49 randomly chosen locations along a water line. Suppose that the lead levels across all the locations on this line are strongly skewed to the right with a mean of $17 \mathrm{ppb}$ and a standard deviation of $14 \mathrm{ppb}$. Assume that the measurements in the sample are independent.

What is the probability that the mean lead level from the sample of 49 measurements $\bar{x}$ is less than $15 \mathrm{ppb}$ ?

Choose 1 answer:
(A) $P(\bar{x}<15) \approx 0.02$
(B) $P(\bar{x}<15) \approx 0.16$
(C) $P(\bar{x}<15) \approx 0.30$
(D) $P(\bar{x}<15) \approx 0.44$
(E) We cannot calculate this probability because the sampling distribution is not normal.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the probability that the sample mean $\bar{x}$ is less than 15 ppb, we use the standard normal distribution (Z-distribution). ‖ ‖ step 2 ⋮ We calculate the Z-score for $\bar{x} = 15$ using the formula $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\mu = 17$ ppb is the population mean, $\sigma = 14$ ppb is the population standard deviation, and $n = 49$ is the sample size. ‖ ‖ step 3 ⋮ Substituting the values into the Z-score formula gives $Z = \frac{15 - 17}{14/\sqrt{49}} = \frac{-2}{2} = -1$. ‖ ‖ step 4 ⋮ We look up the Z-score of -1 in the standard normal distribution table or use a calculator to find the cumulative probability. ‖ ‖ step 5 ⋮ The cumulative probability for Z = -1 is approximately 0.16. This is the probability that the sample mean $\bar{x}$ is less than 15 ppb. ‖ ***B*** ∻Key Concept∻ ⚹Central Limit Theorem⚹ ∻Explanation∻ ⚹The Central Limit Theorem states that the sampling distribution of the sample mean will be approximately normally distributed, regardless of the population distribution, provided the sample size is sufficiently large. In this case, with a sample size of 49, we can assume the sampling distribution of the mean is approximately normal even though the population distribution is skewed.⚹◊What is the formula to calculate the probability that the mean lead level from the sample of 49 measurements is less than 15 ppb?⍭ Generate me a similar question◊

Sia