The mean number of emails adele received during a week was 51.
1 answer:
Given:
The mean number of emails adele received during a week was 51.
The numbers received for the six days were 39, 57, 70, 45, 71, 32.
To find:
The number of emails received on the seventh day.
Solution:
Let x be the number of emails received on the seventh day.
We know that,
Multiply both sides by 7.
Subtract 314 from both sides.
Therefore, she received 43 emails on the seventh day.
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Answer:
0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.
Step-by-step explanation:
To solve this question, we need to use the binomial and the normal probability distributions.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Probability the president will have an IQ of at least 107.5
IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that
This probability is 1 subtracted by the p-value of Z when X = 107.5. So
has a p-value of 0.6915.
1 - 0.6915 = 0.3085
0.3085 probability that the president will have an IQ of at least 107.5.
Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.
First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So
has a p-value of 0.9772.
1 - 0.9772 = 0.0228.
Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:
In which
0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.
What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?
0.3085 probability that the president will have an IQ of at least 107.5.
0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.
Independent events, so we multiply the probabilities.
0.3082*0.0451 = 0.0139
0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.