In an airline reservation system, on entering the flight number, the flight schedule and the flight status are displayed. In thi
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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.
