Angela shared a cab with her friends. When they arrived at their destination, they evenly divided the \$j$jdollar sign, j fare a
1 answer:
Answer:
Step-by-step explanation:
There were three friends and the total fare was $j
Each of the friends pays for their cab fare.
Since Angela also paid a $5 Tip
Total Amount Paid by Angela = Share of cab fare + Tip
We can simplify this further by taking the LCM.
The total amount paid by Angela therefore is :
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Answer:
10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.
99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Mildly obese
Normally distributed with mean 375 minutes and standard deviation 68 minutes. So
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.8962.
So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.
Lean
Normally distributed with mean 522 minutes and standard deviation 106 minutes. So
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.0045.
So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes