Suppose a simple random sample of size n = 1000 is obtained...

Suppose a simple random sample of size n = 1000 is obtained from a population whose (a) Describe the sampling distribution of p^. A. Approximately normal, ?p^ = 0.22 and ?p^ ? 0.0003 B. Approximately normal, ?p^ = 0.22 and ?p^ ? 0.0131 C. Approximately normal, ?p^ = 0.22 and ?p^ ? 0.0002 (b) What is the probability of obtaining x = 260 or more individuals with the characteristic? P(x ? 260) = 0.0011 (Round to four decimal places as needed.) (c) What is the probability of obtaining x = 190 or fewer individuals with the characteristic? P(x ? 190) = (Round to four decimal places as needed.)

Transcript

00:01 Hi, i'm david and i'm here to have you answer your question.

00:03 In the question here, we are going to discuss about the summary discussion on the sample proportion.

00:08 Let me remind you that under certain condition, the sample of proportion p -hart will be approaching to the normal.

00:14 And if we turn the p -hat will manage the mean divided by the standard deviation, we will obtain the standard normal.

00:21 In the question here, i see you answered the part a already, so i see even though you've got it go wrong, the right one must be the p -hart.

00:30 It will follow by the normal when the mean of the p hath equal to the z upon 22 standard division of p hath it will equal to the zp .0131 and for the question p we need to find the probability that x will be corrected equal to the 260.

00:52 To find this probability here i need to convert this one into the p hath by device.

01:00 By everything by n we have n equal to 1000 so we divide everything by the 1000 and if we do it we should get this one equal to the probability on the left tail this one will be in the p heart so we have the p heart quite equal to this will be the z 126 now if you find this probability i will convert the p heart into the z by using this formula and then i will have probability of the z quite equal to the z 26 i will minus the mean 0 .202 over the standard deviation and then we'll get equal to the probability that the x quite equal to 2.

01:45 0 .26 minus 0 .202 divided 0101 equal to the 2 equal to the 2 .05.

01:54 Now to find this probability i need to bring up the z table.

01:58 Let me copy the z table and i will put the table on the right here.

02:06 Notice then the table i bring here enhance the negative g star with the probability on the left tail.

02:11 Right now it's not compatible with this probability yet...

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