Answer to Question #133253 in Statistics and Probability for Pramod

Question

a certain type of storage battery lasts on average 3 yeras with a standard deviation of 0.5 year.assuming that battery life is normally distributed find the probability that a given battery will lasts less than 2.3 years and between 2.5 years and 3.3 years

Answer 1

Solution to 1

P(X<2.3)

Find the z-score of x=2.3

$$Z= (x-\mu) /\sigma$$

$$(2.3 - 3)/0.5 = -1.4$$

P(Z<z) can now be obtained from the standard normal table since it follows z~N(0,1)

P(Z<-1.4) = 0.0808



Answer : 0.0808

Solution to 2

P(2.5<X<3.3)

This is given by P(X<3.3) - P(X<2.5)

Obtaining the z-score of 3.3 and 2.5 yields:

$$(3.3-3)/0.5 = 0.6$$ $$(2.5 - 3)/0.5 = - 1$$

respectively.

Now

$$P(X<3.3)-P(X<2.5)$$ $$= P(Z<0.6) - P(Z<-1)$$

From the standard normal table; P(Z<0.6)= 0.7257

P(Z<-1)= 0.1587

Thus: P(2.5<X<3.3) = 0.7257 - 0.1587

= 0.567

Answer : 0.567