the mean undergraduate cost for tuition, fees, room, and board for four - year institutions was $26,489 for…

the mean undergraduate cost for tuition, fees, room, and board for four - year institutions was $26,489 for a recent academic year. suppose the standard deviation is $3204 and that 36 four - year institutions are randomly selected. find the probability that the sample mean cost for these 36 schools is between $24,000 and $26,000.\nthe answer is 0.1787\naskcuddy\ntrue\nfalse

the mean undergraduate cost for tuition, fees, room, and board for four - year institutions was $26,489 for a recent academic year. suppose the standard deviation is $3204 and that 36 four - year institutions are randomly selected. find the probability that the sample mean cost for these 36 schools is between $24,000 and $26,000.\nthe answer is 0.1787\naskcuddy\ntrue\nfalse

Answer

Explanation:

Step1: Calculate the standard error

The standard error formula is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 3204$ and $n = 36$. So $\sigma_{\bar{x}}=\frac{3204}{\sqrt{36}}=\frac{3204}{6}=534$.

Step2: Calculate the z - scores

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. For $\bar{x}_1 = 24000$, $z_1=\frac{24000 - 26489}{534}=\frac{- 2489}{534}\approx - 4.66$. For $\bar{x}_2 = 26000$, $z_2=\frac{26000 - 26489}{534}=\frac{-489}{534}\approx - 0.92$.

Step3: Find the probabilities

We use the standard normal distribution table. $P(-4.66<Z<-0.92)=P(Z < - 0.92)-P(Z < - 4.66)$. Since $P(Z < - 4.66)\approx0$, and from the standard - normal table $P(Z < - 0.92)=0.1788$. So $P(-4.66<Z<-0.92)\approx0.1788\approx0.1787$.

Answer:

True