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Krystal Ramirez
2024
Question 4, 8.2.11-T
HW Score: 
42.86%,3 of 7 points
Part 4 of 5
Points: 0 of 1
Save
ose a simple random sample of size 
h=200 is obtained from a population whose size is 
N=20,000 and whose population proportion with a specified characteristic is 
p=0.6. lete parts (a) through (c) below.
Not normal because 
n <= 0.05N and 
np(1-p) >= 10.
Approximately normal because 
n <= 0.05N and 
np(1-p) < 10.
Not normal because 
n <= 0.05N and 
np(1-p) < 10.
Approximately normal because 
n <= 0.05N and 
np(1-p) >= 10.
mine the mean of the sampling distribution of 
hat(p).
6 (Round to one decimal place as needed.)
mine the standard deviation of the sampling distribution of 
hat(p).
03464 (Round to six decimal places as fleeded.)
hat is the probability of obtaining 
x=122 or more individuals with the characteristic? That is, what is 
P( hat(p) >= 0.61) ?

0.61)=◻ (Round to four decimal places as needed.)
Get more help
Clear all
Check answe
e

qquad
53.85
Mean: 
◻

◻ Std. Dev.: 052915

Krystal Ramirez\newline20242024\newlineQuestion 44, 88.22.1111-T\newlineHW Score: 42.86%,3 42.86 \%, 3 of 77 points\newlinePart 44 of 55\newlinePoints: 00 of 11\newlineSave\newlineose a simple random sample of size h=200 \mathrm{h}=200 is obtained from a population whose size is N=20,000 \mathrm{N}=20,000 and whose population proportion with a specified characteristic is p=0.6 p=0.6 . lete parts (a) through (c) below.\newlineNot normal because n0.05 N n \leq 0.05 \mathrm{~N} and np(1p)10 n p(1-p) \geq 10 .\newlineApproximately normal because n0.05 N n \leq 0.05 \mathrm{~N} and np(1p)<10 n p(1-p)<10 .\newlineNot normal because n0.05 N n \leq 0.05 \mathrm{~N} and np(1p)<10 n p(1-p)<10 .\newlineApproximately normal because n0.05 N n \leq 0.05 \mathrm{~N} and np(1p)10 n p(1-p) \geq 10 .\newlinemine the mean of the sampling distribution of h=200 \mathrm{h}=200 22.\newline66 (Round to one decimal place as needed.)\newlinemine the standard deviation of the sampling distribution of h=200 \mathrm{h}=200 22.\newline0346403464 (Round to six decimal places as fleeded.)\newlinehat is the probability of obtaining h=200 \mathrm{h}=200 44 or more individuals with the characteristic? That is, what is h=200 \mathrm{h}=200 55 ?\newlineh=200 \mathrm{h}=200 66 (Round to four decimal places as needed.)\newlineGet more help\newlineClear all\newlineCheck answe\newlinee\newlineh=200 \mathrm{h}=200 77\newline5353.8585\newlineMean: h=200 \mathrm{h}=200 88\newlineh=200 \mathrm{h}=200 88 Std. Dev.: 052915052915

Full solution

Q. Krystal Ramirez\newline20242024\newlineQuestion 44, 88.22.1111-T\newlineHW Score: 42.86%,3 42.86 \%, 3 of 77 points\newlinePart 44 of 55\newlinePoints: 00 of 11\newlineSave\newlineose a simple random sample of size h=200 \mathrm{h}=200 is obtained from a population whose size is N=20,000 \mathrm{N}=20,000 and whose population proportion with a specified characteristic is p=0.6 p=0.6 . lete parts (a) through (c) below.\newlineNot normal because n0.05 N n \leq 0.05 \mathrm{~N} and np(1p)10 n p(1-p) \geq 10 .\newlineApproximately normal because n0.05 N n \leq 0.05 \mathrm{~N} and np(1p)<10 n p(1-p)<10 .\newlineNot normal because n0.05 N n \leq 0.05 \mathrm{~N} and np(1p)<10 n p(1-p)<10 .\newlineApproximately normal because n0.05 N n \leq 0.05 \mathrm{~N} and np(1p)10 n p(1-p) \geq 10 .\newlinemine the mean of the sampling distribution of h=200 \mathrm{h}=200 22.\newline66 (Round to one decimal place as needed.)\newlinemine the standard deviation of the sampling distribution of h=200 \mathrm{h}=200 22.\newline0346403464 (Round to six decimal places as fleeded.)\newlinehat is the probability of obtaining h=200 \mathrm{h}=200 44 or more individuals with the characteristic? That is, what is h=200 \mathrm{h}=200 55 ?\newlineh=200 \mathrm{h}=200 66 (Round to four decimal places as needed.)\newlineGet more help\newlineClear all\newlineCheck answe\newlinee\newlineh=200 \mathrm{h}=200 77\newline5353.8585\newlineMean: h=200 \mathrm{h}=200 88\newlineh=200 \mathrm{h}=200 88 Std. Dev.: 052915052915
  1. Calculate Mean: Calculate the mean of the sampling distribution of p^\hat{p} using the formula μp^=p\mu_{\hat{p}} = p.
  2. Calculate Standard Deviation: Calculate the standard deviation of the sampling distribution of p^\hat{p} using the formula σp^=p(1p)n\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}.
  3. Determine Probability: Determine the probability of obtaining x=122x = 122 or more individuals with the characteristic, which translates to p^0.61\hat{p} \geq 0.61. Use the normal approximation to the binomial distribution. First, convert p^\hat{p} to a z-score.
  4. Look up Z-score: Look up z=0.2887z = 0.2887 in the z-table or use a calculator to find the probability that zz is greater than or equal to 00.28872887.

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