Question

A certain type of storage battery lasts, on average 3.0 years with a standard deviation of 0.5 year. Assuming that battery life is normally distributed, answer the following question. a. What is the probability that a randomly selected above types of battery will last less than 2.3 years

Answer 1

Answer:

Probability that a randomly selected above types of battery will last less than 2.3 years = 0.084 .

Step-by-step explanation:

We are given that a certain type of storage battery lasts on average i.e. Mean, $\mu$ = 3.0 years with a Standard deviation, $\sigma$ = 0.5 year.

Let X = randomly selected above types of battery

Also, given battery life is normally distributed so;

                          Z = $\frac{X - \mu}{\sigma}$ ~ N(0,1)

Now, Probability that a randomly selected above types of battery will last less than 2.3 years is given by, P(X < 2.3) .

 P(X < 2.3) = P( $\frac{X - \mu}{\sigma}$ < $\frac{2.3- 3}{0.5}$ ) = P(Z < -1.4) = P(Z > 1.4)

From z% score table we find that P(Z > 1.4) = 0.084 .

Hence, the probability that a randomly selected above types of battery will last less than 2.3 years is 0.084 .

Answer 2

Answer:

Step-by-step explanation:

Assuming that the battery life is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = battery life in years

µ = mean

σ = standard deviation

From the information given,

µ = 3

σ = 0.5

We want to find the probability that a randomly selected above types of battery will last less than 2.3 years. It is expressed as

P(x < 2.3)

For x = 2.3

z = (2.3 - 0.3)/0.5 = - 1.4

Looking at the normal distribution table, the probability corresponding to the z score is 0.08076