You need to calculate the difference in the left-hand areas from a standard normal table (or use z score calculator - Google to find it) for z(4) and z(-2). The area is the probability.
z = (x--µ)/σ where μ=1 and σ=3/sqrt(9) = 1
Please consider a tuto. Take care.
JACQUES D.
tutor
You look up left integrals for higher Z and lower z. The difference in the areas will be the probability.
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09/15/22
Felix K.
After solving it, I found 𝑃 ( − 2 < 𝑋̅ < 4 ) = 0.9973002
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09/16/22
JACQUES D.
tutor
Yes. That is right for -3<z<3. If that is not correct, it could be that sigma of the population is 3^2 = 9 (I don't know why it would be written that way - that's why I assumed it was sigma squared). If that is the case, then sigma of the sample is 3 which would lead to 68% going from -1 to 1 in z.
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09/16/22
Felix K.
thanks
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09/18/22
Felix K.
So after calculating the z-score for z(4) and z(-2). the final answer is P(-1 < Z < 1) ? or just 1?09/15/22