SOLUTION: I am a little stuck from this point A survey found that people keep their television sets an average of 4.8 years. The standard deviation is 0.89 year, and assume that this vari

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Question 1031822: I am a little stuck from this point
A survey found that people keep their television sets an average of 4.8 years. The standard deviation is 0.89 year, and assume that this variable (length of time) is normally distributed. If a person decides to by a new TV set, find the probability that he or she has owned the old set for the following amount of time:Z=X-µ/б
a. Less than 2.5 years
-P(X<2.5) Z=2.5-4.8/0.89
b. Between 3 and 4 years
-P(3
c. More than 4.2 years
-P(X>4.2)
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi,

Using Excel function NORMSDIST

P(X <2.5)= P( z = (2.5-4.8)/0.89) = P(z = -2.584)) = NORMSDIST(-2.5841) = .00493

Let You finish these up ...

P(3 ≤ x ≤ 4) = P(x< 4) - P(x<3)= P( z = (4-4.8)/0.89) - P( z = (3-4.8)/0.89)

P(X > 4.2) = 1 - P(x < 4.2)) = 1 - P( z = (4.2-4.8)/0.89 ) =  

1 - NORMSDIST(-.8989) = 1 - .184 = .816 Or 81.6% 

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