Find the probability that the sample mean is less than $68,680. Round to four decimal places. P(x


Find the probability that the sample mean is less than $68,680. Round to four decimal places.<br>P(x < 68680) =<br>Find the probability that the sample mean is less than $61,120. Round to four decimal places.<br>P(x < 61120)=<br>If you did find a sample mean of MORE THAN $68,680 would you find that unusual? What could you<br>conclude?<br>O It is not unusual to find a sample mean more than $68,680, since the probability is at least 5%.<br>If you find a sample mean more than $68,680, then it may indicate that the population mean<br>has not changed.<br>O It is not unusual to find a sample mean more than $68,680, since the probability is less than<br>5%. If you find a sample mean more than $68,680, then it may indicate that the population<br>mean has not changed.<br>O It is unusual to find a sample mean more than $68,680, since the probability is less than 5%. If<br>you find a sample mean more than $68,680, then it may indicate that the population mean<br>has changed.<br>O It is unusual to find a sample mean more than $68,680, since the probability is at least 5%. If<br>you find a sample mean more than $68,680, then it may indicate that the population mean<br>has changed.<br>

Extracted text: Find the probability that the sample mean is less than $68,680. Round to four decimal places. P(x < 68680)="Find" the="" probability="" that="" the="" sample="" mean="" is="" less="" than="" $61,120.="" round="" to="" four="" decimal="" places.="" p(x="">< 61120)="If" you="" did="" find="" a="" sample="" mean="" of="" more="" than="" $68,680="" would="" you="" find="" that="" unusual?="" what="" could="" you="" conclude?="" o="" it="" is="" not="" unusual="" to="" find="" a="" sample="" mean="" more="" than="" $68,680,="" since="" the="" probability="" is="" at="" least="" 5%.="" if="" you="" find="" a="" sample="" mean="" more="" than="" $68,680,="" then="" it="" may="" indicate="" that="" the="" population="" mean="" has="" not="" changed.="" o="" it="" is="" not="" unusual="" to="" find="" a="" sample="" mean="" more="" than="" $68,680,="" since="" the="" probability="" is="" less="" than="" 5%.="" if="" you="" find="" a="" sample="" mean="" more="" than="" $68,680,="" then="" it="" may="" indicate="" that="" the="" population="" mean="" has="" not="" changed.="" o="" it="" is="" unusual="" to="" find="" a="" sample="" mean="" more="" than="" $68,680,="" since="" the="" probability="" is="" less="" than="" 5%.="" if="" you="" find="" a="" sample="" mean="" more="" than="" $68,680,="" then="" it="" may="" indicate="" that="" the="" population="" mean="" has="" changed.="" o="" it="" is="" unusual="" to="" find="" a="" sample="" mean="" more="" than="" $68,680,="" since="" the="" probability="" is="" at="" least="" 5%.="" if="" you="" find="" a="" sample="" mean="" more="" than="" $68,680,="" then="" it="" may="" indicate="" that="" the="" population="" mean="" has="">
Suppose the mean starting salary for nurses is $67,290 nationally. The standard deviation is<br>approximately $10,478. The starting salary is not normally distributed but it is mound shaped. A<br>sample of 45 starting salaries for nurses is taken.<br>State the random variable.<br>The mean starting salary of a nurse.<br>Starting salary for a nurse.<br>O The standard deviation of starting salaries of nurses.<br>What is the mean of the distribution of all sample means?<br>Hx =<br>What is the standard deviation of the distribution of all sample means? Round to five decimal<br>places.<br>Ox =<br>%3|<br>What is the shape of the sampling distribution of the sample mean? Why?<br>You can say the sampling distribution of the sample mean is normally distributed since the<br>sample size is greater than 30.<br>O You can say the sampling distribution of the sample mean is not normally distributed since<br>the sample size is greater than 30.<br>O You can't say anything about the sampling distribution of the sample mean, since the<br>population of the random variable is not normally distributed and the sample size is greater<br>than 30.<br>

Extracted text: Suppose the mean starting salary for nurses is $67,290 nationally. The standard deviation is approximately $10,478. The starting salary is not normally distributed but it is mound shaped. A sample of 45 starting salaries for nurses is taken. State the random variable. The mean starting salary of a nurse. Starting salary for a nurse. O The standard deviation of starting salaries of nurses. What is the mean of the distribution of all sample means? Hx = What is the standard deviation of the distribution of all sample means? Round to five decimal places. Ox = %3| What is the shape of the sampling distribution of the sample mean? Why? You can say the sampling distribution of the sample mean is normally distributed since the sample size is greater than 30. O You can say the sampling distribution of the sample mean is not normally distributed since the sample size is greater than 30. O You can't say anything about the sampling distribution of the sample mean, since the population of the random variable is not normally distributed and the sample size is greater than 30.

Jun 10, 2022
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