For a recent 10k run, the finishers are normally distributed with mean 58 minutes and standard...
For a recent 10k run, the finishers are normally distributed with mean 63 minutes and standard deviation 13 minutes. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table, Click here to view page 2 of the standard normal distribution table. a. Determine the percentage of finishers with times between 45 and 75 minutes. Approximately % of finishers had times between 45 and 75 minutes. (Round to two decimal places as needed.)...
I have got everything correct except for D. If you could be detailed on how you calculate D with the numbers, that would be great!! For a recent 10k run, the finishers are normally distributed with mean 59 minutes and standard deviation 12 minutes. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. Determine the percentage of finishers...
The times of the finishers in the New York City 10km run are normally distributed with mean of minutes and standard deviation minutes. It is known that 70% of finishers have a finish time greater than 60 minutes. Let denote the finishing time for finishers in this race. Find the standard deviation of the finishing time (). The times of the finishers in the New York City 10km run are normally distributed with mean of 61 minutes and standard deviation o minutes. It...
The times of the finishers in the New York City 10km run are normally distributed with mean of 61 minutes and standard deviation ? minutes. It is known that 70% of finishers have a finish time greater than 60 minutes. Let ?denote the finishing time for finishers in this race. Note: Show your R codes/output. Note: You can use the functions pnorm() or qnorm() in R to help you in solving the following parts. The function pnorm(), compute probabilities from...
Assume the hold time of callers to a cable company is normally distributed with a mean of 20 minutes and a standard deviation of 0.5 minute. Determine the percent of callers who are on hold for more than 20 minutes Click here to view page 1 of the standard normal distrbution table, Click here to view page 2 of the standard romal distribution table, % (Round to two decimal places as needed)
6.92 A variable is normally distributed with mean 0 and standard deviation 4 a. Determine and interpret the quartiles of the variable. b. Obtain and interpret the second decile. c. Find the value that 15% of all possible values of the variable exceed. d. Find the two values that divide the area under the correspond- ing normal curve into a middle area of 0.80 and two outside areas of 0.10. Interpret your answer. 6.95 New York City 10-km Run. As...
Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the probability that a randomly selected thermometer reads between - 1.62 and - 0.91 and draw a sketch of the region. Click to view page 1 of the table. Click to view page 2 of the table. Sketch the region. Choose the correct graph below. O A. OB. Oc. The probability is Click to select your answer(s). Find the indicated...
Assume that the random variable X is normally distributed, with mean 60 and standard deviation - 14. Compute the probability P156X565) Be sure to draw a normal curve the corresponding to the probability shaded Click here to view the standard normal distribution table (page 1) Click here to view the standard normaltaistribution table (page 21 Draw a normal curve with the area corresponding to the probability shaded. Choose the correct graph below ОА ОВ. Ос. OD 5606 The probability P(56<X266)...
Assume the random variable X is normally distributed, with mean = 50 and standard deviation 0 - 9. Find the 12th percentile. The 12th percentile is 0 (Round to two decimal places as needed.) Find the Z-scores that separate the middle 67% of the distribution from the area in the tails of the standard normal distribution Click the loon to view a table of areas under the normal curve. The Z-scores are a (Use a comma to separate answers as...
The mean exam score for 44 male high school students is 20.1 and the population standard deviation is 4.7. The mean exam score for 58 female high school students is 19.6 and the population standard deviation is 4.1. At a = 0.01, can you reject the claim that male and female high school students have equal exam scores? Complete parts (a) through (e). Click here to view page 1 of the standard normal distribution table. Click here to view page...