Question

A report announced that the median sales price of new houses sold one year was $221,000, and the mean sales price was $272,400. Assume that the standard deviation of the prices is $100,000. Complete parts (a) through (d) below. if you select samples of n=2, describe the shape of the sampling distribution of X Choose the correct answer below. A. The sampling distribution is skewed to the right (because population was skewed to the right), but less skewed to the right than the population. OB. The sampling distribution will be approximately normal. C. The sampling distribution is skewed to the left (because population was skewed to the left), but less skewed to the left than the population D. The sampling distribution will be approximately uniform. (b) If you select samples of n = 100, describe the shape of the sampling distribution of X Choose the correct answer below. OA. The sampling distribution will depend on the specific sample and will not have a constant shape. OB. The sampling distribution will be approximately uniform. C. The sampling distribution will be approximately normal. OD. The sampling distribution is skewed to the right, but lessed skew to the right than the population. (c) If you select a random sample of n = 100, what is the probability that the sample mean will be less than $300,000? The probability that the sample mean will be less than $300.000 is 0.9971 (Round to four decimal places as needed.) If you select a random sample of n = 100, what is the probability that the sample mean will be between $275,000 and $295,000? The probability that the sample mean will be be between $275,000 and $295,000 is 0.3855 (Round to four decimal places as needed.)

Answer 1

Sali- Som the samping Given that A Report announced that the median sales price of new house sold one year was 221,000 meam (-4)= 2,72,400 Standard deviation (0) = 1,00,000 5 The Sampling distribution is skewed to the right (because population was skewed to the right than the population) 12 → The Sampling will be approximately normal Q n=100, the sample mean will be less than 3,00,000 meanin) = 2,72,400 Standard deviation (0) = 100000 7100 = 100,000 = 10,000 P(x< 300000) Ų < 3,00,000-272400 10,000 = P(Z < 27,600 10,000 = P(Z < 2.76) By using normal distribution. P(x< 300000) = 0.9971099 P(x< 3,00,000)=10.9971] 10

n=100 the mean will be 2,75,000 and 295000 The probability that the sample mean will be 2,75000 and 29 5000 → PC275000<Z<295000) = P / 27 5000 - 27240042,295000 -272400) 1,00,000 100000 11000000 100000 EP (0.026 <Z < 0.226) = P(220.226) – P (220.026) = 0.5894 -0.5104 = 10.07901