if the random variable z is a standard normal score what is p(-2.00<_z<_+2.00)? how did you find the probabilty please show and or explain

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The probability for (-2<z<2) = 1-(pr(z<= -2) + pr(z>=2))

This reads 'the probability z is between -2 and 2 equals 1 minus the probability that z is less than or equal to -2 and z is greater than or equal to 2'.

This is because the probability that z is less than or equal to -2, the probability that z is between -2 and 2 and the probability that z is greater than or equal to 2 all sum to one - there are no other possibilities for z.

Now, the normal distribution is symmetric about z=0 so

pr(z>=2) = pr(z<= -2)

This symmetry is very useful when using tables which generally only give you pr(z<=a) for some value a.

Now that we know this we then have that

pr(-2<z<2) = 1-(pr(z<= -2) + pr(z>=2))

                  = 1- 2*pr(z<=-2)

                  = 1-2*Phi(-2)

where Phi is the cumulative distribution function of the normal distribution given in the table. If you look at the value for -2 in the table the probability is 0.023 so

pr(-2<z<2) = 1-2*0.023 = 1- 0.046 = 0.954

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You integrate the Normal distribution between -2 and 2. This can't be done in closed form so you would need to use a computer or look up tables.

However, +/-2 standard deviations is a common thing to look at so you should know that this is roughly probability 0.95 (0.954).

On a computer you would do 1-Phi(-2)*2 = 0.954 where Phi is the cdf of the standard Normal distribution ie the probabililty in each tail is roughly 0.025

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