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ISSION TO ENTER FOR WORK ORDERS! C Secure https//wwwwebassignnetweb/student/Assignment-Responses/submittdep 15492200 o ea pants Previous Answers Mistastars SE025. variation both in the actual glucose level and in the blood test that measures the level. concerned that she may suffer from gestational dabetes (high blood glucose levels during pregnancy). There is Sheila's doctor is A patient is classified as having gestational diabetes ir the glucose level is above 140 milligrams per decilter (mga) one hour after a sugary drink is ingested shella's measured gucose level one hour after ingesting the sugary drink varies according to the Normal distribution with 12 mg/dl and o a) if a single glucose measuremert is made, what is the probability that Sheila is diagnosed as having gestational dabetes? (Round your answer to four decimal places.) (b) measurements are made instead on 2 separate days and the mean result is compared with the criterion 140 mg/d, what is the probability that sheila is diagnosed as having gestational diabetes (Round answer to four decimal places.) points 5E027 level and in the blood test that measures the level Sheila's doctor is that suffer from gestational dabetes (high blood glucose levels duuringpregnaney) There is variation both in the actual gaucese level Meur aner angeseng classified as having gestational dabetes if the glucose level is above 140 milugrams per decanter (mga) one hour an a is gested Sheilas measured the sugary drink varies according to the Normal distribution H 123 mg/dl and a for Sheilas glucose level dstritution? (Round your answer to one decimal place.) Assam e a sample n picture of the distribution of the r statiste under mehypothes the t distraution critical values talle and yeur picture the values of test that 23. Draw a the use would lead to of the null hypothesis at the level for a two sided alternative PARK a: 208 336 8787 CABLE ONE FAx 208 342 3630

Explanation / Answer

1.

a.

P(X>140)
= P(Z > 140-128 /10)
= P(Z> 1.2)
=1-.8849
=.1151

b.

P(X>140)
= P(Z > 140-128 /(10/sqrt(2))
= P(Z> 1.697)
The P-Value is 1-.9554 = .0446

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