This question requires the use of a standard normal table. Let z be a standard normal random variable, find P(z > -1.17). Enter your answer rounded to 4 decimal places.
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- Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(csZs1.1)=D0.8518 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(1.07Let z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(-csZsc)=0.9625 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places. ?Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(Zsc) =0.1685 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.You may need to use the appropriate appendix table to answer this question. Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.) (d) The area to the right of z is 0.1271. (e) The area to the left of z is 0.6700. (f) The area to the right of z is 0.3300.Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(1.18Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c . =P ( Z ≤ C ) = 0.7123 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.You may need to use the appropriate appendix table to answer this question. Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.) (a) The area to the left of z is 0.2119. (b) The area between −z and z is 0.9398. (c) The area between −z and z is 0.2052. (d) The area to the left of z is 0.9951. (e) The area to the right of z is 0.5793.Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(Zsc)=0.8888 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(cYou may need to use the appropriate appendix table or technology to answer this question. z is a standard normal random variable. Find P(z ≥ 2.12).What is the probability of randomly selecting a z-score less than z = –1.25 from a normal distribution? Question options: a. 0.2112 b. 0.1056 c. 0.8944 d. 0.3944SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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