Questions: Question Lexie averages 149 points per bowling game with a standard deviation of 14 points. Suppose Lexie's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X ~ N(149,14). If necessary, round to three decimal places. Provide your answer below: Suppose Lexie scores 186 points in the game on Tuesday. The z-score when x=186 is . The mean is . This z-score tells you that x=186 is standard deviations to the right of the mean.

Question
Lexie averages 149 points per bowling game with a standard deviation of 14 points. Suppose Lexie's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X ~ N(149,14).

If necessary, round to three decimal places.

Provide your answer below:

Suppose Lexie scores 186 points in the game on Tuesday. The z-score when x=186 is . The mean is .

This z-score tells you that x=186 is  standard deviations to the right of the mean.
Transcript text: Question Lexie averages 149 points per bowling game with a standard deviation of 14 points. Suppose Lexie's points per bowling game are normally distributed. Let $X=$ the number of points per bowling game. Then $X \sim N(149,14)$. If necessary, round to three decimal places. Provide your answer below: Suppose Lexie scores 186 points in the game on Tuesday. The $z$-score when $x=186$ is $\square$ . The mean is $\square$ . This $z$-score tells you that $x=186$ is $\square$ standard deviations to the right of the mean. FEEDBACK MORE INSTRUCTION SUBMIT
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Solution

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Solution Steps

Step 1: Calculate the Z-Score

To find the z-score for Lexie's score of x=186 x = 186 , we use the formula:

z=Xμσ z = \frac{X - \mu}{\sigma}

Substituting the values:

z=18614914=37142.643 z = \frac{186 - 149}{14} = \frac{37}{14} \approx 2.643

Step 2: Identify the Mean

The mean of Lexie's bowling scores is given as:

μ=149 \mu = 149

Step 3: Interpret the Z-Score

The calculated z-score indicates how many standard deviations the score of x=186 x = 186 is from the mean. Since z2.643 z \approx 2.643 , we conclude that:

x=186 is approximately 2.643 standard deviations to the right of the mean. x = 186 \text{ is approximately } 2.643 \text{ standard deviations to the right of the mean.}

Final Answer

The z-score when x=186 x = 186 is 2.643 2.643 . The mean is 149 149 . This z-score tells you that x=186 x = 186 is 2.643 2.643 standard deviations to the right of the mean.

z=2.643,μ=149 \boxed{z = 2.643, \mu = 149}

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