Let z be a standard normal random variable with mean ? = 0 and standard deviation ? = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.) a. P(z < 1) = b. P(z > 1.15) = c. P(−2.31 < z < 2.31) = d. to the left of 1.8 e.Let z be a standard normal random variable with mean ? = 0 and standard deviation ? = 1. Find the percentile. (Round your answer to two decimal places.) z0.20or the 80th percentile

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19-12-2022
Mathematics

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Let z be a standard normal random variable with mean ? = 0 and standard deviation ? = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.) a. P(z < 1) = b. P(z > 1.15) = c. P(−2.31 < z < 2.31) = d. to the left of 1.8 e.Let z be a standard normal random variable with mean ? = 0 and standard deviation ? = 1. Find the percentile. (Round your answer to two decimal places.) z0.20or the 80th percentile

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Please Answer with Explanation

qwarrow qwarrow · Answer 1 · 2022-12-22T10:57:01-05:00

From the standard normal distribution table, the probability value is 0.1357.

The variable Z denotes a standardized form of a ‘normal distribution’ which follows normal with a mean(μ)=0, SD(σ)=1.

A probability distribution that is symmetric about the mean is the normal distribution, sometimes referred to as the Gaussian distribution. It demonstrates that data that are close to the mean occur more frequently than data that are far from the mean.

The normal distribution appears as a "bell curve" on a graph.

We have,

P(Z ＞ –1.1) = 1 - P(Z ＜ -1.1)

From the standard normal distribution table, the probability value corresponding to row = -1.1 and column = 0.00 is 0.1357.

Therefore, P(Z＜-1.1) = 0.1357

Thus,

P(Z ＞ –1.1) = 1 - P(Z ＜ -1.1)

P(Z ＞ –1.1) = 1 - 0.1357

P(Z ＞ –1.1) = 0.8643

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