Question

let z be a standard normal random variable. find z1 such that p(-2.3<z<z1)= 0.1046

Answer #1

Here, we are having

p(-2.3<z<z1)= 0.1046

Thus, P(z < z1) - P(z < -2.3) = 0.1046

Now, P(z < -2.3) = 0.0107 (This is obtained through, z table. P(z < -2.3) is the area of normal curve towards left of -2.3)

Thus,

P(z < z1) - 0.0107 = 0.1046

Thus, P(z < z1) = 0.1046 + 0.0107 = 0.1153

Now, in order to find z1 we need to find value corresponding to probability 0.1153

Thus, referring to standard normal distribution z table

we get, **z1 = -1.2**

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calculation.
A. 1.28
B. -1.28
C. 0.255
D. -0.255
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four decimal places.)
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(Round all answers to two decimal places)
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? =
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? =
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Please be as detailed as possible, thank
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