Students of a large university spend an average of $7 a day on lunch. The standard...

Students of a large university spend an average of $7 a day on lunch. The standard deviation of the expenditure is $2. A simple random sample of 25 students is taken. What is the probability that the sample mean will be at least $4? Jason spent $15 on his lunch. Explain, in terms of standard deviation, why his expenditure is not usual. Explain what information is given on a z table. For example, if a student calculated a z value of 2.77, what is the four-digit number on the z table that corresponds with that value? What exactly is that 4-digit number telling us? Explain why we use z formulas. Why don't we just leave the data alone? Why do we convert? must show work

Expert Solution

orchestra answered 1 year ago

Students of a large university spend an average of $5 a day on lunch. The standard...

Students of a large university spend an average of $5 a day on lunch. The standard deviation of the expenditure is $3. A simple random sample of 36 students is taken. What are the expected value, standard deviation, and shape of the sampling distribution of the sample mean? What is the probability that the sample mean will be at least $4? What is the probability that the sample mean will be at least $5.90?

Students of a large university spend an average of $6 a day on lunch. The standard...

Students of a large university spend an average of $6 a day on lunch. The standard deviation of the expenditure is $2. A simple random sample of 81 students is taken. 1. What is the probability that the sample mean will be at least $5.25? 2. What is the probability that the sample mean will be at least $6.50? 3. What is the range of money spent by people who fall within one standard deviation of the mean? 4. Kelsey...

A small university knows the average amount that its students spend on lunch each day. The...

A small university knows the average amount that its students spend on lunch each day. The amount spent on lunches for the population of 500 students is not highly skewed and has a mean of $8 and a standard deviation of $2. Suppose simple random sample of 49 students is taken, what is the probability that the sample mean for the sample of 49 students will be between $7.50 and $8.50?

At a large university, the students have an average creditcarddebt of $2500,with a standard deviation of...

At a large university, the students have an average creditcarddebt of $2500,with a standard deviation of $1200. If we consider all of the possible random samples of 100students at this university, 95% of thesamples should have means between what two numbers? [HINT-use the 68-95-99.7Rule table in conjunction with the values for the sampling distribution model] A)$100 and $2620 (B)$300 and $4900 (C)$2140 and $2860 (D)$2260 and $2740

A survey of college students reported that they spend an average of $9.50 a day on...

A survey of college students reported that they spend an average of $9.50 a day on dinner with a standard deviation of $3. What is the probability that 100 randomly selected college students will spend less than $10.00 on average for dinner? Round your answer to 4 decimal places.

A large Midwestern university is interested in estimating the mean time that students spend at the...

A large Midwestern university is interested in estimating the mean time that students spend at the student recreation center per week. A previous study indicated that the standard deviation in time is about 25 minutes per week. If the officials wish to estimate the mean time within ± 4 minutes with a 90 percent confidence, what should the sample size be? 106 Can't be determined without the sample mean. 105 105.685

A sample of university students has an average GPA of 2.78 with a standard deviation of...

A sample of university students has an average GPA of 2.78 with a standard deviation of 0.45. If GPA is normally distributed, what percentage of the students has GPAs….. More than 2.3? Less than 3.0? Between 2.00 and 2.5?

Based on all student records at Camford University, students spend an average of 6.00 hours per...

Based on all student records at Camford University, students spend an average of 6.00 hours per week playing organized sports. The population’s standard deviation is 3.40 hours per week. Based on a sample of 49 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates. a) Compute the standard error of the sample mean. (Round your answer to 2 decimal places.) Standard error b) What is the chance HLI will find a sample...

The average amount you spend on a lunch during the week is not known. Based on...

The average amount you spend on a lunch during the week is not known. Based on past experience, you are willing to assume that the standard deviation is $2.10. You take a random sample of 28 lunches, and apply the central limit theorem. Round your answer to 2 decimal places. Fill in the blank: The 68–95–99.7 rule says that the probability is about 0.95 that is within $ ____ of the population mean mu.

A population of 1,000 students spends an average of $10.50 a day on dinner. The standard...

A population of 1,000 students spends an average of $10.50 a day on dinner. The standard deviation of the expenditure is $3. A simple random sample of 64 students is taken. a. What are the expected value, standard deviation, and shape of the sampling distribution of the sample mean? b. What is the probability that these 64 students will spend an average of more than $11 per person? c. What is the probability that these 64 students will spend an...

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