SOLUTION: Q.2 By using Normal Distribution, solve the following: a) A certain type of battery lasts, on average, 3.0 years with a standard deviation of 0.5 years. Assuming that the batter

Algebra ->  Probability-and-statistics -> SOLUTION: Q.2 By using Normal Distribution, solve the following: a) A certain type of battery lasts, on average, 3.0 years with a standard deviation of 0.5 years. Assuming that the batter      Log On


   

Question 1137508: Q.2 By using Normal Distribution, solve the following:
a) A certain type of battery lasts, on average, 3.0 years with a standard deviation of 0.5 years. Assuming that the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years.
b) An electrical firm manufactures light bulbs that have a life, before burn-out, that is normally distributed with mean equal to 800 hours and a standard deviation of 40 hours. Find the probability that a bulb burns between 778 and 834 hours.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Part A

mu = 3 = mean
sigma = 0.5 = standard deviation

Convert the raw score x = 2.3 to its corresponding z score
z = (x-mu)/sigma
z = (2.3-3)/0.5
z = -1.40

Use this table (or similar) to find that P(Z < -1.40) = 0.0808
To find this result, turn to page 1 of that PDF, go to the row that starts with -1.4 and the column that has 0.00 at the top. The row and column intersect at 0.0808


The probability is approximately 0.0808
which is roughly 8.08% in percent form
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Part B

mu = 800
sigma = 40

Convert the raw score x = 778 to its corresponding z score
z = (x-mu)/sigma
z = (778-800)/40
z = -0.55

Repeat for x = 834
z = (x-mu)/sigma
z = (834-800)/40
z = 0.85

So, P(778 < X < 834) is the same as P(-0.55 < Z < 0.85)

We will use this formula here
P(A < Z < B) = P(Z < B) - P(Z < A)
to help compute the area we want between the two z scores

A = -0.55
B = 0.85
P(A < Z < B) = P(Z < B) - P(Z < A)
P(-0.55 < Z < 0.85) = P(Z < 0.85) - P(Z < -0.55)
P(-0.55 < Z < 0.85) = 0.8023 - 0.2912 use the same table as done in part A
P(-0.55 < Z < 0.85) = 0.5111


The approximate probability is 0.5111
which in percent form would be roughly 51.11%

Here is a handy calculator to help check your work