When Sophie goes bowling, her scores are normally distributed with a mean of 110 and a standard deviation of 10. What percentage of the games that Sophie bowls does she score higher than 129, to the nearest tenth?
Q. When Sophie goes bowling, her scores are normally distributed with a mean of 110 and a standard deviation of 10. What percentage of the games that Sophie bowls does she score higher than 129, to the nearest tenth?
Identify mean and standard deviation: Identify the mean (μ) and standard deviation (σ) of Sophie's bowling scores.The mean (μ) is given as 110, and the standard deviation (σ) is given as 10.
Calculate z-score for Sophie's score: Calculate the z-score for Sophie's score of 129. The z-score formula is z=(X−μ)/σ, where X is the score in question. For Sophie's score of 129, the z-score would be z=(129−110)/10.
Perform z-score calculation: Perform the calculation for the z-score.z=10129−110z=1019z=1.9
Find probability using standard normal distribution: Use a standard normal distribution table or calculator to find the probability of scoring less than a z-score of 1.9.The table or calculator will give us the area to the left of the z-score.
Subtract area to find higher scores percentage: Subtract the area found in Step 4 from 1 to find the area to the right of the z-score, which represents the percentage of games Sophie scores higher than 129. If the area to the left of z=1.9 is P, then the area to the right is 1−P.
Look up area in z-table: Look up the area to the left of z=1.9 in the z-table or use a calculator to find P. Assuming a standard normal distribution, the area to the left of z=1.9 is approximately 0.9713.
Calculate area for higher scores: Calculate the area to the right of z=1.9, which is the percentage of games Sophie scores higher than 129.Area to the right = 1−0.9713Area to the right ≈0.0287
Convert area to percentage: Convert the area to the right into a percentage.Percentage = 0.0287×100Percentage ≈2.87%
Round percentage: Round the percentage to the nearest tenth.Rounded percentage = 2.9%
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