Empirical research on stock market data for two consecutive trading days indicates that 60% of the stocks that went up on the first day also went up on the second day. Yesterday, 600 stocks went up. (a) Find the mean of p̂, where p̂ gives the proportion of the 600 stocks that went up yesterday that will go up today. (b) Find the standard deviation of p̂ (c) Compute an approximation for P (p̂ < 0.64), which is the probability that fewer than 64% of the stocks that went up yesterday will go up again today. Round your answer to four decimal places.

Empirical research on stock market data for two consecutive trading days indicates that 60% of the stocks that went up on the first day also went up on the second day. Yesterday, 600 stocks went up. (a) Find the mean of p̂, where p̂ gives the proportion of the 600 stocks that went up yesterday that will go up today. (b) Find the standard deviation of p̂ (c) Compute an approximation for P (p̂ < 0.64), which is the probability that fewer than 64% of the stocks that went up yesterday will go up again today. Round your answer to four decimal places.

Name: Empirical research on stock market data for two consecutive trading da
Uploaded: 2024-03-20T07:06:41.000Z
Channel: Bartleby
Description: Empirical research on stock market data for two consecutive trading days indicates that 60% of the stocks that went up on the first day also went up on the second day. Yesterday, 600 stocks went up.(a) Find the mean of p̂, where p̂ gives th

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

See similar textbooks

Similar questions

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 188000 dollars. Assume the standard deviation is 34000 dollars. Suppose you take a simple random sample of 61 graduates.Find the probability that a single randomly selected policy has a mean value between 171893 and 175375.6 dollars.P(171893 < X < 175375.6) = (Enter your answers as numbers accurate to 4 decimal places.)Find the probability that a random sample of size n=61n=61 has a mean value between 171893 and 175375.6 dollars.P(171893 < M < 175375.6) = (Enter your answers as numbers accurate to 4 decimal places.)
Assume that the returns from an asset are normally distributed. The average annual return for this asset over a specific period was 17.1 percent and the standard deviation of those returns in this period was 41.7 percent. a. What is the approximate probability that your money will double in value in a single year? (Do not round intermediate calculations and enter your answer as a percent rounded to 2 decimal places, e.g., 32.16.) b. What about triple in value? (Do not round intermediate calculations and enter your answer as a percent rounded to 6 decimal places, e.g., 32.161616.)
S1
Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 152000 dollars. Assume the standard deviation is 31000 dollars. Suppose you take a simple random sample of 58 graduates. Find the probability that a single randomly selected policy has a mean value between 137753.3 and 150371.8 dollars. P(137753.3
A psychologist obtains a random sample of 30 mothers in the first trimester of their pregnancy. The mothers are asked to play Mozart in the house at least 45 minutes each day until they give birth. We know that IQs are normally distributed with a mean of 100. A few years after their births the children were given an IQ test and the IQs of the 30 children in the study resulted in a sample mean of 104.2. Assume the population standard deviation for IQ of children is 15. Is there evidence that the children who listen to Mozart for at least 45 minutes each day during pregnancy have higher IQs at the 0.05 level of significance? Step 1: Define the parameter & setup the testStep 2: State the Level of SignificanceStep 3: Find the value of the Test StatisticsStep 4: Find P-Value OR Find Critical ValueStep 5: State Conclusion and why
The percent of fat calories that a person consumes each day is normally distributed with a mean of about 35 and a standard deviation of about ten. Suppose that 9 individuals are randomly chosen. Let X = average percent of fat calories. For the group of 9, find the probability that the average percent of fat calories consumed is more than six. (Round your answer to four decimal places.)
The undergraduate grade point averages (UGPA) of students taking an admissions test in a recent year can be approximated by a normal distribution, as shown in the figure. (a) What is the minimum UGPA that wouldstill place a student in the top5 % of UGPAs? (b) Between what two values does the middle 50 % of the UGPAs lie?
Suppose a life insurance company sells a $170,000 1-year term life insurance policy to a 20-year-old female for $170. According to the National Vital Statistics Report, 58(21), the probability that the female survives the year is 0.999544. The expected value of this policy to the insurance company is $92.48. What is the standard deviation of the value of the life insurance policy? Why is the value so high?
Fifteen items or less: The number of customers in line at a supermarket express checkout counter is a random variable with the following probability distribution. X 0 1 2 3 4 5 P(x) 0.05 0.25 0.30 0.25 0.10 0.05 Send data to Excel
A pizza company advertises that it puts 0.5 pounds of real mozzarella cheese on its medium pizzas. Suppose that the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of 0.5 pounds and a standard deviation of 0.005 pounds. (Let y = the amount of cheese on a medium pizza. Round the answers to three decimal places.) (a) What is the probability that the amount of cheese on a medium pizza is between 0.505 and 0.510 pounds?P(0.505 ≤ y ≤ 0.510) = (b) What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations?P(y > ? + 2?) = (c) What is the probability that four randomly selected medium pizzas all have at least 0.495 pounds of cheese?
Women's heights are normally distributed with a mean given by u = 63.6 in. and a standard deviation given by o = 2.5 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 62.6 in. Enter a number correct to 4 decimal places: (b) If 52 women are randomly selected, find the probability that they will have a mean height less than 62.6 in. Enter a number correct to 4 decimal places:
Suppose the mean cholesterol levels of women age 45-59 is 5.2 mmol/l and the standard deviation is 0.7 mmol/l. Assume that cholesterol levels are normally distributed. Find the probability that a woman age 45-59 has a cholesterol level above 6.1 mmol/l (considered a high level). Round to four decimal places.P(x > 6.1) = Suppose doctors decide to test the woman’s cholesterol level again and average the two values. Find the probability that this woman’s mean cholesterol level for the two tests is above 6.1 mmol/l. Round to four decimal places.P(x̄ > 6.1) = Suppose doctors being very conservative decide to test the woman’s cholesterol level a third time and average the three values. Find the probability that this woman’s mean cholesterol level for the three tests is above 6.1 mmol/l. Round to four decimal places.P(x̄ > 6.1) =