In a recent study on world happiness, participants were asked to evaluate their current lives on a scale from 0 to 10, where 0 represents the worst possible life and 10 represents the best possible life. The responses were normally distributed, with a mean of 6.1 and a standard deviation of 2.5. Answer parts (a)–(d) below. (a) Find the probability that a randomly selected study participant’s response was less than 4. The probability that a randomly selected study participant’s response was less than 4 is ______. (Round to four decimal places as needed.) (b) Find the probability that a randomly selected study participant’s response was between 4 and 6. The probability that a randomly selected study participant’s response was between 4 and 6 is ______. (Round to four decimal places as needed.) (c) Find the probability that a randomly selected study participant’s response was more than 8. The probability that a randomly selected study participant’s response was more than 8 is ______. (Round to four decimal places as needed.) (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below. - A. The events in parts (a) and (c) are unusual because their probabilities are less than 0.05. - B. The event in part (a) is unusual because its probability is less than 0.05. - C. The events in parts (b), (a), and (c) are unusual because all of their probabilities are less than 0.05. - D. There are no unusual events because all the probabilities are greater than 0.05.
It is given that the responses are normally distributed with mean 6.1 and standard deviation 2.5.
Denote the random variable X as the response of a randomly selected person.
That is, µ=6.1, σ = 2.5.
Thus, X~ N (6.1, 2.5).
a)
The probability that a randomly selected study participant's response was less than 4 is calculated as follows:
P(X<4) = P(Z<((4-6.1)/2.5) (by standardizing)
= P(Z<-0.84) From the standard normal table, P(Z<-0.84) = 0.2005
= 0.2005
Thus, probability that a randomly selected study participant's response was less than 4 is 0.2005.
The probability value obtained using the Excel formula, =NORM.DIST(4,6.1,2.5,1) will be 0.2005.
b)
The probability that a randomly selected study participant's response was between 4 and 6 is calculated as follows:
P(4<X<6) = P(Z<((6-6.1)/2.5) -P(Z<((4-6.1)/2.5) (by standardizing)
= P(Z<-0.04)- P(Z<-0.84)
From the standard normal table, P(Z<-0.04)= 0.4840 and P(Z<-0.84) = 0.2005
= 0.4840 – 0.2005
= 0.2835
Thus, the probability that a randomly selected study participant's response was between 4 and 6 is 0.2835.
The probability value obtained using the Excel formula, =NORM.DIST(6,6.1,2.5,1)- NORM.DIST(4,6.1,2.5,1) will be 0.2836.
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