The formula for the area of the shaded region on the diagram is: Area of the circle - Area of the square The area of the circle
1 answer:
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Answer:
Option (1) is the correct option.
Step-by-step explanation:
From the given frequency table,
Option (1).
Tossing an even number is more likely than tossing an odd number.
Frequencies of tossing even numbers 2, 4, 6 = 10, 10, 9
Frequencies of tossing odd numbers 1, 3, 5 = 8, 6, 7
Which shows frequencies of tossing even numbers are more.
Therefore, Option (1) is correct.
Option (2)
The number 2 will be the next number tossed.
Since it is unpredictable, therefore, this option is incorrect.
Option (3)
Number 4 is half as likely to be tossed as the number 3.
Frequency of tossing number 3 = 6
Frequency of tossing 4 = 10
Frequency of tossing 3 is not the half of tossing 4.
Therefore, Option (3) is incorrect.
Option (4)
Probability of tossing the number 2 =
=
=
=
Therefore, option (4) is incorrect.
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