Chapter 13, Problem 42AP

a

Summary Introduction

Interpretation:

Proportion of bulb lasting more than 2 years.

Concept Introduction:

Mean is the average value of the data given. It is usually the middle value which represents the whole data. It is calculated by summing up all values divided by umber of values.

Answer to Problem 42AP

The probability of bulbs lasting more than two years is 0.48%.

Explanation of Solution

Given information:

Time Period = 2 year

Mean = 2.670

The proportion of bulbs lasting more than two years using the following formula:

  $P\left[T\ge t\right]=e^{-\mu }$

Here, t is 2 years and $\lambda$ is 2.670. Substitute the values in the equation and calculate the proportion of bulbs lasting more than two years as shown below:

  $P\left[T\ge t\right]=e^{-\mu }$

  $=e^{-2.670\times 2}$

  $=e^{-5.34}$

  $={2.178}^{-5.34}$

  $=0.0048$

Thus, the probability of bulbs lasting more than two years is 0.48%.

b

Summary Introduction

Interpretation:

Probability that a bulb chosen fails in first-three months of operation.

Concept Introduction:

Mean is the average value of the data given. It is usually the middle value which represents the whole data. It is calculated by summing up all values divided by umber of values.

Answer to Problem 42AP

Probability that a bulb chosen at random fails is 48.69%.

Explanation of Solution

Given information:

Time = 3 months

Mean = 2.670

The probability that a bulb chosen at random fails using the following formula:

  $P\left[T\ge t\right]=1-e^{-\mu }$

Here, t is 3/12 years and $\lambda$ is 2.670. Substitute the values in the equation and calculate the probability that a bulb chosen at random fails as shown below:

  $P\left[T<\frac{3}{12}\right]=1-e^{-\mu }$

  $=1-e^{-2.670\times \frac{3}{12}}$

  $=1-e^{-2.89}$

  $=1-{2.718}^{-2.89}$

  $=0.486$

Thus, the probability that a bulb chosen at random fails is 48.69%.

c

Summary Introduction

Interpretation:

Probability that a bulb lasted for 10 years fails in next three months.

Concept Introduction:

Mean is the average value of the data given. It is usually the middle value which represents the whole data. It is calculated by summing up all values divided by umber of values.

Answer to Problem 42AP

The probability of bulbs that has lasted for 10 years fails in the next three months of operation is1.9102E+13%.

Explanation of Solution

Given information:

Time = 10 years

Mean = 2.670

The probability that a bulb that has lasted for 10 years fails in the next three months of operation using the following formula:

  $P\left[t<T<t-s\left|T>t\right|\right]=\frac{P\left(T<s\right)}{P\left(T>t\right)}\mathrm{.......}\left(1\right)$

Here, t is 10 years and s is 3/12. Substitute the values in the equation and calculate the proportion of bulb that has lasted for 10 years fails in the next three months of operation as shown below:

  $P\left[T<\frac{3}{12}\right]=1-e^{-\mu }$

  $=1-e^{-1.670\times \frac{3}{12}}$

  $=1-e^{-2.89}$

  $=1-{2.718}^{-2.89}$

  $=0.486$

Thus, P (T < s) is 0.486.

  $\begin{array}{l}P\left[T>10\right]=e^{-\mu }\\ =e^{-2.670\times 10}\\ =e^{-26.7}\\ ={2.718}^{-26.7}\end{array}$

Thus, P (T > 10) is $={2.718}^{-26.7}$ .

Substitute the values in the equation (1) and calculate as shown below:

  $P\left[10<T<10-3/12\left|T>10\right|\right]$

  $=\frac{P\left(T<s\right)}{P\left(T>t\right)}$

  $=\frac{0.0486}{{2.718}^{-26.7}}$

  $=1.9E+11$

Thus, the probability of bulbs that has lasted for 10 years fails in the next three months of operation is 1.9102E+13%.