Question

QUESTION 10 NeverReady batteries has engineered a newer, longer-lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. You randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, What is the probability of getting a random sample of 30 batteries in which the sample mean lifetime is 16.7 hours or less? (Round to 3 decimal places)

Answer 1

Here we used the concept of Central limit Theorem from tht of Large sample theory and the concept of standard Normal distribution to get the required answer.

Given :- The company claims this battery has an average life span of 17 hours with a Standard deviation of 0.8 hours. A pandom sample of 30 batteries is selected and found out that the Sample life span is 16-7 hours Now what is the prob. of getting a sample of 30 batteries in which the sample mean is 16.7 houps or less ?

Note :- The central limit the distribution of theorem sample states that means distribution apppoximates to a normal as the sample size keeps on increasing The required sample size is generally equal to op greater than 30 Let {xn} be a sequence of ind pandom variables with E(X) = and u(x) =82200 Define Sn = 3 xk Then by Central Limit Theorem, Sn-E(Sn) d (0,1) Vu(sn) 2 vn (Xn- l d N (0,1) le. tends to in disth N(0,1) Let X;= lifespan of a battery from the given population, i=1,2...n (x) = M = 17 hours, vor (xn) = 6² = (0.8)² = 0.64 hours Suppose Sn = 2 2 2 X be the sum of the pondomine chosen 30 batteries sample e S-2X [no] K - 1

Then by eut (contral Limit Theorem) Vão ( 780-17) + (0) Now, Probability [getting a sample of 30 batteres in which the sample mean is a 16-7 -17 1-17 =P[716.1] - (Rv **). (1270 0.8

= (-2-05396) = 1- & (2-05396) 1-0-9198 ſcaleulaning from standard normal 396 table = 10.0202 ... The required probability 18 0-020 2 .

Rounding off to 3 decimal places we get the answer as 0.02

Now how to calculate from a standard normal distribution table.

The table is:

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Look for the value 2.0 from the left most column. Now for the value .05 from the right most row. The respective row and column for the particular values meet at the cell with the value 0.9798

For any sort of doubts ask in the comment section below.

A ? would be really appreciable.