A fuel pump at a gasoline station doesn't always dispense the exact amount displayed on the meter. When the meter reads $1.000 \mathrm{~L}$, the amount of fuel a certain pump dispenses is normally distributed with a mean of $1 \mathrm{~L}$ and standard deviation of $0.05 \mathrm{~L}$. Let $X=$ the amount dispensed in a random trial when the meter reads $1.000 \mathrm{~L}$.

Find $P(X

Q: A fuel pump at a gasoline station doesn't always dispense the exact amount displayed on the meter. When the meter reads $1.000 \mathrm{~L}$, the amount of fuel a certain pump dispenses is normally distributed with a mean of $1 \mathrm{~L}$ and standard deviation of $0.05 \mathrm{~L}$. Let $X=$ the amount dispensed in a random trial when the meter reads $1.000 \mathrm{~L}$. Find $P(X<1)$. You may round your answer to two decimal places.

∻Solution by Steps∻ ‖ step 1 ⋮ To find $P(X < 1)$ for a normally distributed variable, we need to standardize the variable using the Z-score formula. ‖ ‖ step 2 ⋮ The Z-score is calculated by the formula $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value of interest, $\mu$ is the mean, and $\sigma$ is the standard deviation. ‖ ‖ step 3 ⋮ Substituting the given values into the Z-score formula gives us $Z = \frac{1 - 1}{0.05} = \frac{0}{0.05} = 0$. ‖ ‖ step 4 ⋮ However, we need the Z-score for $X < 1$, so we calculate $Z = \frac{1 - 1}{0.05} = \frac{0}{0.05} = 0$. This is the Z-score for the exact mean, so we need to look at the Z-score for a value just less than 1. ‖ ‖ step 5 ⋮ Since the Z-score for $X = 1$ is 0, and we want $P(X < 1)$, we look up the Z-score for a value just less than 0 in the standard normal distribution table. ‖ ‖ step 6 ⋮ The Z-score for values less than 0 in the standard normal distribution table corresponds to a probability of 0.5. ‖ ‖ step 7 ⋮ Therefore, $P(X < 1) = 0.5$. ‖ ∻Answer∻ ⚹ $P(X

Question

A fuel pump at a gasoline station doesn't always dispense the exact amount displayed on the meter. When the meter reads $1.000 \mathrm{~L}$, the amount of fuel a certain pump dispenses is normally distributed with a mean of $1 \mathrm{~L}$ and standard deviation of $0.05 \mathrm{~L}$. Let $X=$ the amount dispensed in a random trial when the meter reads $1.000 \mathrm{~L}$.

Find $P(X<1)$.
You may round your answer to two decimal places.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find $P(X < 1)$ for a normally distributed variable, we need to standardize the variable using the Z-score formula. ‖ ‖ step 2 ⋮ The Z-score is calculated by the formula $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value of interest, $\mu$ is the mean, and $\sigma$ is the standard deviation. ‖ ‖ step 3 ⋮ Substituting the given values into the Z-score formula gives us $Z = \frac{1 - 1}{0.05} = \frac{0}{0.05} = 0$. ‖ ‖ step 4 ⋮ However, we need the Z-score for $X < 1$, so we calculate $Z = \frac{1 - 1}{0.05} = \frac{0}{0.05} = 0$. This is the Z-score for the exact mean, so we need to look at the Z-score for a value just less than 1. ‖ ‖ step 5 ⋮ Since the Z-score for $X = 1$ is 0, and we want $P(X < 1)$, we look up the Z-score for a value just less than 0 in the standard normal distribution table. ‖ ‖ step 6 ⋮ The Z-score for values less than 0 in the standard normal distribution table corresponds to a probability of 0.5. ‖ ‖ step 7 ⋮ Therefore, $P(X < 1) = 0.5$. ‖ ∻Answer∻ ⚹ $P(X < 1) = 0.5$ ⚹ ∻Key Concept∻ ⚹Standardizing a Normal Distribution⚹ ∻Explanation∻ ⚹To find probabilities related to normal distributions, we standardize the variable to a Z-score and use the standard normal distribution table to find the corresponding probability◊What is the formula to calculate the z-score for the given scenario of a normal distribution⍭ Generate me a similar question◊⚹

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