O O According to a certain organization, adults worked an average of 1,760 hours last year. Assume the population standard deviation is 380 hours and that a random sample of 50 adults was selected. Complete parts a through e below. CEITES a. Calculate the standard error of the mean. 0₂-0 (Round to two decimal places as needed.) b. What is the probability that the sample mean will be more than 1,770 hours? P(X>1,770) = (Round to four decimal places as needed.). c. What is the probability that the sample mean will be between 1,720 and 1,750 hours? P(1,720 sxs 1,750) = (Round to four decimal places as needed.) d. Would a sample mean of 1,787 hours support the claim made by the organization? Select the correct choice below and fill in the answer box within your choice.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.5: Comparing Sets Of Data
Problem 3CYU
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Question
According to a certain organization, adults worked an average of 1,760 hours last year. Assume the population standard deviation is 380 hours and that a random sample of 50 adults was selected. Complete parts a through e below.
a. Calculate the standard error of the mean.
(Round to two decimal places as needed.)
b. What is the probability that the sample mean will be more than 1,770 hours?
P(x> 1,770) =
(Round to four decimal places as needed.)
c. What is the probability that the sample mean will be between 1,720 and 1,750 hours?
P (1,720 ≤x≤ 1,750) =
(Round to four decimal places as needed.)
d. Would a sample mean of 1,787 hours support the claim made by the organization? Select the correct choice below and fill in the answer box within your choice.
(Round to four decimal places as needed.)
OA. The probability P (x21,787) =
OB. The probability P(x21,787) =
esc
hours ≤x≤ hours
(Round to one decimal place as needed.)
e. Identify the symmetrical interval that includes 95% of the sample means if the true population mean is 1,760 hours.
1
:0
F1
2
This contradicts the claim.
F2
This supports the claim.
#
3
20
F3
$
4
DO
F4
%
5
F5
MacBook Air
6
F6
7
F7
* CO
► 11
8
F8
1
9
F9
0
Time Remaining: 01:50:01
F10
7
F11
+ 11
Next
43) F12
G
delete
Transcribed Image Text:According to a certain organization, adults worked an average of 1,760 hours last year. Assume the population standard deviation is 380 hours and that a random sample of 50 adults was selected. Complete parts a through e below. a. Calculate the standard error of the mean. (Round to two decimal places as needed.) b. What is the probability that the sample mean will be more than 1,770 hours? P(x> 1,770) = (Round to four decimal places as needed.) c. What is the probability that the sample mean will be between 1,720 and 1,750 hours? P (1,720 ≤x≤ 1,750) = (Round to four decimal places as needed.) d. Would a sample mean of 1,787 hours support the claim made by the organization? Select the correct choice below and fill in the answer box within your choice. (Round to four decimal places as needed.) OA. The probability P (x21,787) = OB. The probability P(x21,787) = esc hours ≤x≤ hours (Round to one decimal place as needed.) e. Identify the symmetrical interval that includes 95% of the sample means if the true population mean is 1,760 hours. 1 :0 F1 2 This contradicts the claim. F2 This supports the claim. # 3 20 F3 $ 4 DO F4 % 5 F5 MacBook Air 6 F6 7 F7 * CO ► 11 8 F8 1 9 F9 0 Time Remaining: 01:50:01 F10 7 F11 + 11 Next 43) F12 G delete
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1->
According to a certain organization, adults worked an average of 1,760 hours last year. Assume the population standard deviation is 380 hours and that a random sample of 50 adults was selected. Complete parts a through e below.
O
a. Calculate the standard error of the mean.
0-= 53.74
(Round to two decimal places as needed.)
b. What is the probability that the sample mean will be more than 1,770 hours?
P(x>1,770) = 9.4246
(Round to four decimal places as needed.)
c. What is the probability that the sample mean will be between 1,720 and 1,750 hours?
P(1,720 ≤x≤ 1,750) = 0.1951
(Round to four decimal places as needed.)
esc
d. Would a sample mean of 1,787 hours support the claim made by the organization? Select the correct choice below and fill in the answer box within your choice.
(Round to four decimal places as needed.)
OA. The probability P(x21,787) =
OB. The probability P(x21,787) =
1
e. Identify the symmetrical interval that includes 95% of the sample means if the true population mean is 1,760 hours.
hours ≤x≤ hours
(Round to one decimal place as needed.)
F1
2
W
This contradicts the claim.
F2
This supports the claim.
#
3
F
F3
54
$
4
R
F4
%
5
H
F5
T
10
MacBook Air
A
6
F6
Y
&
7
F7
* CO
8
► 11
F8
9
F9
с
0
Time Remaining: 01:07:03
F10
4)
F11
A
+ 11
Next
F12
x
G
delete
Transcribed Image Text:1-> According to a certain organization, adults worked an average of 1,760 hours last year. Assume the population standard deviation is 380 hours and that a random sample of 50 adults was selected. Complete parts a through e below. O a. Calculate the standard error of the mean. 0-= 53.74 (Round to two decimal places as needed.) b. What is the probability that the sample mean will be more than 1,770 hours? P(x>1,770) = 9.4246 (Round to four decimal places as needed.) c. What is the probability that the sample mean will be between 1,720 and 1,750 hours? P(1,720 ≤x≤ 1,750) = 0.1951 (Round to four decimal places as needed.) esc d. Would a sample mean of 1,787 hours support the claim made by the organization? Select the correct choice below and fill in the answer box within your choice. (Round to four decimal places as needed.) OA. The probability P(x21,787) = OB. The probability P(x21,787) = 1 e. Identify the symmetrical interval that includes 95% of the sample means if the true population mean is 1,760 hours. hours ≤x≤ hours (Round to one decimal place as needed.) F1 2 W This contradicts the claim. F2 This supports the claim. # 3 F F3 54 $ 4 R F4 % 5 H F5 T 10 MacBook Air A 6 F6 Y & 7 F7 * CO 8 ► 11 F8 9 F9 с 0 Time Remaining: 01:07:03 F10 4) F11 A + 11 Next F12 x G delete
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