<p>Can somebody help me with this Stat problem please? It would be greatly appreciated :)</p>
<p>Sam and Janet Evening want to estimate the average dollar amount of the orders filled by their South Pacific Catering Company. They obtain their estimate by selecting a SRS of 49 orders. Sam and Janet don't know it, but their orders are normally distributed with mu=$120 and sigma=$21.
Within what range of values does the x(bar) have a 95 percent chance of falling? </p>
<p>Thanks.</p>
<p>Mean of mean of sample population: $120; standard deviation of the mean of sample population: $21/sqr(49)=$3</p>
<p>InvNorm(.025,120,3)
InvNorm(.975,120,3)</p>
<p>You get the range if you’re talking about middle 95%.</p>
<p>I’m not positive, but I think this question may be meant to use confidence intervals…we just started this chapter so I’m not positive, but I’m not sure invnorm will give the answer the teacher wants. And I’m not entirely sure how to work with confidence intervals yet…maybe after class today I can help, lol.</p>
<p>It may be meant to use confidence intervals, but jerrry’s method still answers the question. Unfortunately, a lot of topics that could be (and really SHOULD be) connected in AP statistics aren’t.</p>
<p>mu sub x-bar = mu = 120
sigma sub x-bar = sigma/sqrt(n) = 21/sqrt(49)= 21/7 = 3
The empirical rule (aka 68-95-99.7 rule) states that 68% of observations will fall within 1 standard deviation (the sigma sub x-bar), 95% will fall within 2s, 99.7% will fall within 3s.</p>
<p>So for 95%, you have:
mean ± 2s
= mu sub x-bar ± 2sigma sub x-bar
= 120 ± 2*3
= 120 ± 6.
Your range would be 114 to 126.</p>
<p>if it’s confidence value, then zhangm94 method is correct, but it’s
mu sub x-bar ± 1.9599 sigma sub x-bar</p>