34.-12 points My Notes Ask Your Teache Assume that the weight of hamsters is Nor

Question

34.-12 points My Notes Ask Your Teache Assume that the weight of hamsters is Normally distributed with = 1 lbs and = 0.25 lbs. Suppose I randomly select 6 hamsters. (a) What is the shape of the sampling distribution of the sample mean? O Unknown O Approximately Normal O Exactly Normal (b) What is P1.13)? (Use 3 decimal places) Submit Answer Save Progress Practice Another Version 35. +-3 points My Notes Ask Your Teache Sheila's doctor is concerned that she may suffer from gestational diabetes (high blood glucose levels during pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. A patient is classified as having gestational diabetes if her glucose level is above 135 milligrams per deciliter (mg/dl) one hour after a sugary drink. Sheila's measured glucose level one hour after the sugary drink varies according to the Normal distribution with = 120 mg/dl and = 11 mg/dl. Let X = Sheila's measured glucose level one hour after a sugary drink (a) P(X > 135) Suppose measurements are made on 4 separate days and the mean result is compared with the criterion 135 (Use 3 decimal places) ri^/kil. (b) P(X > 135) = (c) What sample mean blood glucose level is higher than 95% of all other sample mean blood glucose levels? Hint (Use 3 decimal places) this requires a backward Normal calculation. (Use 2 decimal places)

Explanation / Answer

34)a)m exactly normal as population is normal

b)std error =std deviation/(n)1/2 =0.25/(6)1/2 =0.1021

therefore P(Xbar<1.13)=P(Z<(1.13-1)/0.1021)=P(Z<1.2737)=0.899

35)a )P(X>135)=P(Z>(135-120)/11)=P(Z>1.3636)=0.086

b) std error =11/(4)1/2 =5.5

P(Xbar>135) =P(Z>(135-120)/5.5)=P(Z>2.7273)=0.003

c)for 95 percentile ; z=1.645

therefore corresponding sample mean =120+1.645*5.5=129.05

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