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2025-03-24JST06:51:25 Someone%20is%20described%20as%20%E2%80%9Crisk%20loving%E2%80%9D%20if%20the%20maximum%20amount%20he%20or%20she%20is%20willing%20to%20pay%20to%20play%20is%20greater%20than%20$50,%20the%20game%E2%80%99s%20expected%20value.%20Suppose%20in%20this%201,536-person%20sample,%2020%20people%20are%20risk%20loving.%20Let%20p%20denote%20the%20proportion%20of%20the%20adult%20population%20aged%2022%20to%2055%20who%20are%20risk%20loving%20and%201%20%E2%80%93%20p,%20the%20proportion%20of%20the%20same%20population%20who%20are%20not%20risk%20loving.%20Use%20the%20sample%20results%20to%20estimate%20the%20proportion%20p.%20what%20is%20the%2090%%20confidence%20interval?Someone%20is%20described%20as%20%E2%80%9Crisk%20loving%E2%80%9D%20if%20the%20maximum%20amount%20he%20or%20she%20is%20willing%20to%20pay%20to%20play%20is%20greater%20than%20$50,%20the%20game%E2%80%99s%20expected%20value.%20Suppose%20in%20this%201,536-person%20sample,%2020%20people%20are%20risk%20loving.%20Let%20p%20denote%20the%20proportion%20of%20the%20adult%20population%20aged%2022%20to%2055%20who%20are%20risk%20loving%20and%201%20%E2%80%93%20p,%20the%20proportion%20of%20the%20same%20population%20who%20are%20not%20risk%20loving.%20Use%20the%20sample%20results%20to%20estimate%20the%20proportion%20p.%20what%20is%20the%2090%%20confidence%20interval?
2025-03-24JST06:51:22 Someone is described as “risk loving” if the maximum amount he or she is willing to pay to play is greater than $50, the game’s expected value. Suppose in this 1,536-person sample, 20 people are risk loving. Let p denote the proportion of the adult population aged 22 to 55 who are risk loving and 1 – p, the proportion of the same population who are not risk loving. Use the sample results to estimate the proportion p. what is the standard deviation?Someone is described as “risk loving” if the maximum amount he or she is willing to pay to play is greater than $50, the game’s expected value. Suppose in this 1,536-person sample, 20 people are risk loving. Let p denote the proportion of the adult population aged 22 to 55 who are risk loving and 1 – p, the proportion of the same population who are not risk loving. Use the sample results to estimate the proportion p. what is the standard deviation?
2025-03-24JST06:51:19 Someone is described as “risk loving” if the maximum amount he or she is willing to pay to play is greater than $50, the game’s expected value. Suppose in this 1,536-person sample, 20 people are risk loving. Let p denote the proportion of the adult population aged 22 to 55 who are risk loving and 1 – p, the proportion of the same population who are not risk loving. Use the sample results to estimate the proportion p.Someone is described as “risk loving” if the maximum amount he or she is willing to pay to play is greater than $50, the game’s expected value. Suppose in this 1,536-person sample, 20 people are risk loving. Let p denote the proportion of the adult population aged 22 to 55 who are risk loving and 1 – p, the proportion of the same population who are not risk loving. Use the sample results to estimate the proportion p.
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