Which of the operating systems listed below was the last to be released
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Based on the mean of the subscribers, the sample size, and the standard deviation, the probability of females being at most 0.25 is 0.2202.
The probability that they are between 0.22 and 0.28 is 0.2703.
The probability that they are within 0.03 of the population proportion is 0.3566.
<h3>What is the probability that they are at most 0.25?</h3>Using Excel, we shall assume that the distribution is normally distributed.
We can therefore use the NORM.DIST function:
=NORM.DIST(0.25,0.3,0.0648,TRUE)
= 0.2202
<h3 /><h3>What is the probability that they are between 0.22 and 0.28?</h3>=NORM.DIST(0.28,0.3,0.0648,TRUE) - NORM.DIST(0.22,0.3,0.0648,TRUE)
= 0.2703
<h3>What is the probability that they are within 0.03 of the population proportion?</h3>X high = 0.30 + 0.03
= 0.33
X low = 0.30 - 0.03
= 0.27
= NORM.DIST(0.33,0.3,0.0648,TRUE) - NORM.DIST(0.27,0.3,0.0648,TRUE)
= 0.3566.
Find out more on probability at brainly.com/question/1846009.
Answer:
Following are the solution to this question:
Please find the attachment of this code.
Explanation:
In this code, an email_list method is declared, that accepts a domains parameter, and in the next step, an empty list and two for loop are defined, in which the first for loop is used for a count parameter value and in the second loop is use the append method to add the given value, and use the return method to print its value, and use the print method to call the "email_list" method.
