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Let \( Z \sim N(0,1) \). Accurate to 4 decimal places, evaluate \( P(-2.58
Ask by Estrada Coles. in the United States
Mar 09,2025

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The probability \( P(-2.58 < Z < 2.65) \) is approximately 0.991.

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To find \( P(-2.58 < Z < 2.65) \) when \( Z \) follows a standard normal distribution \( N(0,1) \), we will use the cumulative distribution function (CDF) values for the corresponding z-scores. First, look up the CDF values: - \( P(Z < 2.65) \) can be found in standard normal distribution tables or using software/calculator, which gives approximately 0.9962. - \( P(Z < -2.58) \) also can be found similarly and gives approximately 0.0049. Now, we can find the probability: \[ P(-2.58 < Z < 2.65) = P(Z < 2.65) - P(Z < -2.58) = 0.9962 - 0.0049 = 0.9913. \] Rounding this to four decimal places, we get \( P(-2.58 < Z < 2.65) \approx 0.9913 \). Thus, the answer is **0.991**.

Related Questions

Question 12(Mulliple Choice Warth 5 points) \[ (04.06 \mathrm{HC}) \] A researcher wants to test the claim that the proportion of juniors who watch television regularly is greater than the proportion of seniors who watch television regularly She finds that 56 of 70 randomly selected juniors and 47 of 85 randomly selected seniors report watching television regularly. Construct \( 95 \% \) confidence intervals for each population proportion. Which of the statemente gives the correct outcome of the research or's tert of the dalim? The \( 95 \% \) confidence interval for juniors is (706, 894), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is (721, 879), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( 706,894 ), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( \( 721, .879 \) ), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors.

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