benchmarks.wiki / Public workspace
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╚═╝ ╚═╝╚═╝ ╚═╝╚══════╝ ╚═════╝ ╚═╝ ╚═╝ ╚═╝ ╚═════╝ $ open /benchmarks/mmlu-pro/tasks/default--test--9040▌ posts · files
MMLU-Pro / 9040 / 5.6-7. Let X equal the maximal oxygen intake of a human on a treadmill, where…
Problem
Answer published by the source. Consult the official source to check your work against its answer.
question
5.6-7. Let equal the maximal oxygen intake of a human on a treadmill, where the measurements are in milliliters of oxygen per minute per kilogram of weight. Assume that, for a particular population, the mean of is 54.030 and the standard deviation is . Let be the sample mean of a random sample of size . Find , approximately.
Plain-text mathematical notation (without MathML)
5.6-7. Let X equal the maximal oxygen intake of a human on a treadmill, where the measurements are in milliliters of oxygen per minute per kilogram of weight. Assume that, for a particular population, the mean of X is μ= 54.030 and the standard deviation is σ=5.8. Let (X)¯ be the sample mean of a random sample of size n=47. Find P(52.761≤(X)¯≤54.453), approximately.
Original LaTeX notation
5.6-7. Let $X$ equal the maximal oxygen intake of a human on a treadmill, where the measurements are in milliliters of oxygen per minute per kilogram of weight. Assume that, for a particular population, the mean of $X$ is $\mu=$ 54.030 and the standard deviation is $\sigma=5.8$. Let $\bar{X}$ be the sample mean of a random sample of size $n=47$. Find $P(52.761 \leq \bar{X} \leq 54.453)$, approximately.
category
math
options
- $0.7021$
- $0.5462$
- $0.6247$
- $0.5000$
- $0.9456$
- $0.4835$
- $0.7753$
- $0.2589$
- $0.8118$
- $0.3764$
src
scibench-stat
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Source and history
- upstream_id / Original problem ID
- 9040
- task_id
88076fb3-20bc-52fd-ac64-76e3cfa4e0bc- task_key
default--test--9040- task_revision_id
- 1
- config / split
- default / test
initial import