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Use the following data. It has been previously established that for a certain type of AA battery (when newly produced), the voltages are distributed normally with $\mu=1.50 \mathrm{V}$ and $\sigma=0.05 \mathrm{V}$.

What percent of the batteries have voltages between $1.52 \mathrm{V}$ and 1.58 V?

Intro Stats / AP Statistics

Chapter 22

Introduction to Statistics

Section 4

Normal Distributions

Descriptive Statistics

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(II) A $12.0-\mathrm{V}$ b…

Welcome to enumerate in the current problem. We have the voltages of aa batteries, mhm of a batteries. And the variable is distributed with mean this 1.50 and Sigma 0.05 goals. Now we have to find the probability of what percentage of the batteries will lie between 1.52 To 1.58. Now we we cannot find this using uh empirical table because for that we will need this exactly what six that we can find. But if it is particular points like this, which are not on particular sigma levels, then we To a standardization. So 1.5 to -1.5 divided by 0.05, Less than it was two. X -1.5 divided by 0.05. Less than equal to I will write down Who? -1.5 divided by 0.05. So if we simplify that, we will get 0.02 Divided by 0.05. Less than said, less than over here. It will be zero 08 divided by 0.0 Fight, correct? This can be written as probability two by five. Less than said. Less than eight by fight. Which is equals two, probability 0.4. Less than zero, less than one point six. Well now, if we bring the standard normal table over here, treat me this, but I will just it used the size. So this will be Probability of zed less than 1.6 minus probability. Zeb less than zero food. What do we mean by that? See if this is a distribution and we want the values off. See Z equals to zero four, correct. 0.4. So we will read that Valium I'm here. So you see it's 0.6554. So it will be this area will be 0.6554. So let us write 0.6554 and 1.6 will be somewhere over here. So that will have a bigger area, correct? So if you go and see 1.6 and 0.945 to 0.9452 therefore, if we subtract, we get eight and then nine and then eight and then 892 So this is the probability. So the percentage will be 28.98%, which can be rounded up to 23%. So 23% of the Batteries will be having a lifetime, uh will be having a four days between 1.52 and 1.5. It so I hope you could understand this. Let me know if you have any questions.

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