In: Statistics and Probability
1. An expensive watch is powered by a 3-volt lithium battery expected to last
five years. Suppose the life of the battery has a standard deviation of 0.3 year and is normally distributed.
a. Determine the probability that the watch's battery will last longer than 5.4 years.
b. Calculate the probability that the watch's battery will last more than 4.45 years.
c. Compute the length-of-life value for which 15% of the watch's batteries last longer.
2. Doggie Nuggets Inc. (DNI) sells large bags of dog food to warehouse clubs. DNI uses an automatic filling process to fill the bags. Weights of the filled bags are approximately normally distributed with a mean of 56 kilograms and a standard deviation of 0.86 kilograms. Complete parts a through d below.
a. What is the probability that a filled bag will weigh less than 55.2 kilograms?
1)
a)
Here, μ = 5, σ = 0.3 and x = 5.4. We need to compute P(X >= 5.4). The corresponding z-value is calculated using Central Limit Theorem
z = (x - μ)/σ
z = (5.4 - 5)/0.3 = 1.33
Therefore,
P(X >= 5.4) = P(z <= (5.4 - 5)/0.3)
= P(z >= 1.33)
= 1 - 0.9082 = 0.0918
b)
Here, μ = 5, σ = 0.3 and x = 4.45. We need to compute P(X >= 4.45). The corresponding z-value is calculated using Central Limit Theorem
z = (x - μ)/σ
z = (4.45 - 5)/0.3 = -1.83
Therefore,
P(X >= 4.45) = P(z <= (4.45 - 5)/0.3)
= P(z >= -1.83)
= 1 - 0.0336 = 0.9664
c)
z value at 15% = 1.04
z = (x - mean)/s
1.04 = (x - 5)/0.3
x = 0.3 *1.04 + 5
x = 5.31
2)'
Here, μ = 56, σ = 0.86 and x = 55.2. We need to compute P(X <=
55.2). The corresponding z-value is calculated using Central Limit
Theorem
z = (x - μ)/σ
z = (55.2 - 56)/0.86 = -0.93
Therefore,
P(X <= 55.2) = P(z <= (55.2 - 56)/0.86)
= P(z <= -0.93)
= 0.1762