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The mean life of a television set is 123 months with a standard deviation of 19 months. If a sample of 55 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by greater than 1.2 months? I just need help getting from 1 - P (-0.47 < z < 0.47) to the answer. In my lesson it says the next step is 0.3192 but I have NO CLUE where that number comes from. Please help me! Thank you!

User Kevtrout
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1 Answer

4 votes

Answer:

1- P(-0.47 < z < 0.47) = 0.63836

Explanation:

Here, we want to get the answer to;

1-P ( -0.47 < z < 0.47)

Mathematically, we are going to use the standard normal distribution table for this

Let us try to evaluate;

P(-0.47 < z < 0.47)

Mathematically, that will be;

P (z < 0.47) - P(z < -0.47)

where P (z < -0.47) = P( z > 0.47)

so we have

P(-0.47 < z < 0.47) = P(z < 0.47) - P(z > 0.47)

So we have this using the standard normal distribution table as follows;

0.68082 - 0.31818 = 0.36164

Now, we can insert this in the earlier expression of;

1- P(-0.47 < z < 0.47)

= 1-0.36164 = 0.63836

User Tobia
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