The length of time a bat stays in a cave has a normal distribution with ? = 95 days and a standard deviation of ?=5.7 days. Suppose that we measure 445 different bats for time in a cave. Let X be the random variable representing the mean number of days and let Xtot be the random variable representing the sum of the days in a cave of the 445 examined bats.
a) About what proportion of bats stay between 90 and 100 days in a
cave?
b) About proportion of bats stay fewer than 85 days in a
cave?
c) About how many of the 445 bats stayed fewer than 85 days in a
cave? (nearest integer)
d) About how many of the 445 bats stayed between 90 and 100 days in
a cave?(nearest integer)
e) What is the standard deviation of the distribution of
X?
f) What is the standard deviation of the distribution of
Xtot?
g) What is the probability that 94.5 < X <
95.5?
h) What is the probability that 42,200 < Xtot < 42,800?
f) Copy your R script for the above into the text box here.
USE R CODE IF POSSIBLE // NOT MANDATORY
a) About what proportion of bats stay between 90 and 100 days in a cave?
R code to get proportion of bats stay between 90 and 100 days in a cave is,
mean = 95
std.dev = 5.7
n = 445
std.err = std.dev / sqrt(n) # Standard error of the mean
# a.
z.upper = (100 - mean)/std.dev
z.lower = (90 - mean)/std.dev
ans.a = pnorm(z.upper) - pnorm(z.lower)
The output of the code is 0.6196182
b) About proportion of bats stay fewer than 85 days in a cave?
R code to get proportion of bats stay fewer than 85 days in a cave is,
# b.
z = (85 - mean)/std.dev
ans.b = pnorm(z)
The output of the code is 0.0396822
c) About how many of the 445 bats stayed fewer than 85 days in a
cave? (nearest integer)
Number of bats stayed fewer than 85 days in a cave is
ans.c = round(n * pnorm(z))
The output of the code is 18.
d) About how many of the 445 bats stayed between 90 and 100 days in
a cave?(nearest integer)
Number of bats stayed between 90 and 100 days in a cave is
ans.d = round(n * (pnorm(z.upper) - pnorm(z.lower)))
The output of the code is 276.
e) What is the standard deviation of the distribution of X?
std.err = std.dev / sqrt(n) # Standard error of the mean
The output of the code is 0.2702059
f) What is the standard deviation of the distribution of Xtot?
std.tot = sqrt(n*std.dev^2) #standard deviation of total
The output of the code is 120.2416
g) What is the probability that 94.5 < X < 95.5?
z.upper = (95.5 - mean)/std.err
z.lower = (94.5 - mean)/std.err
ans.g = pnorm(z.upper) - pnorm(z.lower)
The output of the code is 0.9357499
h) What is the probability that 42,200 < Xtot < 42,800?
z.upper = (42800 - n* mean)/std.tot
z.lower = (42200 - n * mean)/std.tot
ans.h = pnorm(z.upper) - pnorm(z.lower)
The output of the code is 0.7335958
f) Copy your R script for the above into the text box here.
All R codes are given below with screenshot.
mean = 95
std.dev = 5.7
n = 445
# a.
z.upper = (100 - mean)/std.dev
z.lower = (90 - mean)/std.dev
ans.a = pnorm(z.upper) - pnorm(z.lower)
# b.
z = (85 - mean)/std.dev
ans.b = pnorm(z)
# c.
ans.c = round(n * pnorm(z))
# d.
ans.d = round(n * (pnorm(z.upper) - pnorm(z.lower)))
#e.
std.err = std.dev / sqrt(n) # Standard error of the mean
#f.
std.tot = sqrt(n*std.dev^2) #standard deviation of total
# g.
z.upper = (95.5 - mean)/std.err
z.lower = (94.5 - mean)/std.err
ans.g = pnorm(z.upper) - pnorm(z.lower)
# h.
z.upper = (42800 - n* mean)/std.tot
z.lower = (42200 - n * mean)/std.tot
ans.h = pnorm(z.upper) - pnorm(z.lower)
The length of time a bat stays in a cave has a normal distribution with ?...
The distribution of the length of waiting time makes the passport approach the normal distribution with an average of 280 seconds and the standard deviation of 90 seconds. It is known that 75% of all passport owners make it online. A random sample of 50 people was taken. Determine the chance of the proportion of people making passports online between 55% and 65%.
Assume you have a normal distribution representing the likelihood of project completion times. The mean of this distribution is 14, and the standard deviation is 4. The probability of completing the project in 15 or fewer days is: a. 0.59 b. 0.27 c. 0.93 d. 0.75
Name: Wit: SAMPLING DISTRIBUTION EXERCISE The average length of a hospital stay in general or community hospitals in the United States is 6 days Assume that the population standard deviation is 7.7 days. A simple random sample of 49 patients is drawn from this population. 1. What is the sampling distribution of the sample mean? 2. What is the probability that the average length of stay for this group of patients will be no more thana week? 3. What is...
The random variable x has a normal distribution with standard deviation 2525. It is known that the probability that x exceeds 159159 is .90. Find the mean μ of the probability distribution.
The random variable x has a normal distribution with standard deviation 24.It is known that the probability that x exceeds 170 is .90. Find the mean μ of the probability distribution.
(3) Suppose the length of stay in a chronic disease hospital of a certain type of patient has the mean of 60 days with a standard deviation of 15 days. It is reasonably to assume an approximately normal distribution of the lengths of stay (a) If one patient is selected from this group at random, what the probability that this patient will have a length of stay between 50 and 90 days? at is selected from this group at random,...
NORMAL PROBABILITY DISTRIBUTION Medicine: Blood Glucose: A person's blood glucose level and diabetes are closely related. Let x be a random variable measured in milligrams of glucose per deciliter (1/10 of a liter) of blood. After a 12-hour fast, the random variable x will have a distribution that is approximately normal with mean= 85 and standard deviation= 25. Note: After 50 years of age, both the mean and standard deviation tend to increase. What is the probability that, for an...
Assume the time to complete a task has a normal distribution with mean 20 min. and standard deviation 4 min. Find the proportion of times that are BETWEEN 15 and 25 minutes? Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 7 is entered as 7.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38 | | Assume the time to complete a task has a normal distribution with mean 20 min....
Lemonade cup vending machines at JJ Café pour lemonade into cups. The amount of lemonade dispensed has approximately a normal distribution with μ= 7.7 ounces and a standard deviation of σ= 0.41 ounces. Suppose that we randomly select 150 lemonade cups. Let X be the random variable representing the mean amount of lemonade in ounces and let Xtot be the random variable representing the sum of the amounts of ounces of lemonade in the 150 selected cups. e) What is...
Let X be a random variable with a normal distribution having a mean of 30 and a known standard deviation of 16. What is the probability that X is greater than 50? A- 0.1056 B- 0.6057 C- 0.3944 D- 0.8944