Company XYZ know that replacement times for the DVD players it produces are normally distributed with a mean of 7.8 years and a standard deviation of 1.4 years. Find the probability that a randomly selected DVD player will have a replacement time less than 3.5 years? P(X < 3.5 years) = Enter your answer accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. If the company wants to provide a warranty so that only 4.1% of the DVD players will be replaced before the warranty expires, what is the time length of the warranty? warranty = years Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Solution :
Given that ,
mean =
= 7.8
standard deviation =
= 1.4
P(x < 3.5 )
= P[(x -
) /
< ( 3.5 - 7.8 ) / 1.4 ]
= P(z < -3.071 )
Using z table,
= 0.0011
Probability = 0.0011
b.
The z - distribution of the 4.1% is,
P(Z < z) = 4.1%
= P(Z < z ) = 0.041
= P(Z < -1.739 ) = 0.041
z = -1.739
Using z-score formula,
x = z *
+
x = -1.739 * 1.4 + 7.8
x = 5.3654
x = 5.4
Answer : x = 5.4
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