Homework Chapter 7. This assignment is due at the beginning of class on Tuesday,
ID: 3121526 • Letter: H
Question
Homework Chapter 7. This assignment is due at the beginning of class on Tuesday, March 28. Name 1. Sheila's doctor is concerned that she may suffer from gestational diabetes (high blood glucose levels during pregnancy). There is variation both in actual glucose level and in the blood test that measures the level. A patient is classified as having gestational diabetes if the glucose level is above 140 milligrams per deciliter (mg/dl) one hour after consuming a sugary drink. Sheila's measured glucose level one hour after the sugary drink varies according to the Normal distribution with 125 mg/dl and a 10 mg/dl a) If a single glucose measurement is made, what is the probability that Sheila is diagnosed as having gestational diabetes? Hint: Find P(x 140, wherexis Sheila's glucose level. b) If the measurements are made on 4 separate days and the mean result is compared with the criterion 140 mg/dl, what is the probability that Sheila is diagnosed as having gestational diabetes? Hint: Find P(i 140), where Ris Sheila's mean glucose level for the 4 days. 2. A machine manufactures parts whose diameters vary according to the Normal distribution with hu 40.150 mi meters and a standard deviation of o 0.003 mm. An inspector measures a random sample of 4 parts. Find the probability that the average diameter of these 4 parts is less than 40,148 mm.Explanation / Answer
Qn No. 1. Let x denote Sheila’s glucose level.
Given that x is normally distributed with mean = 125 mg/dl and
Standard deviation = 10 mg/dl.
Thus, it is given x follows normal distribution, N( 125,100).
a) We are required to find P(x> 140)
For that we have to standardize x to Z. Then Z =( x- )/
Z = (x- 125)/ 10
Therefore, P(x> 140) = P((x- 125)/ 10 > (140- 125) /10)
= P( Z > 1.5)
From normal distribution table, for the area under the normal curve,
for Z= 1.5, the area from 0 to 1.5 is 0.4332.
To get the required probability, we have to subtract this area from 0.5.
Therefore, P(x > 140) = P(Z> 1.5)
= 0.5-0.4332
= 0.0668
b) To find P(x > 140)
Let x1, x2, x3 and x4 denote the glucose levels of 4 days and let x be their mean.
Each of this x1, x2, x3 and x4 follows normal distribution with mean = 125 mg/dl and Standard deviation = 10 mg/dl.
Therefore, by the property of normal distribution,
x will follow normal distribution with mean = 125 mg/dl and Standard deviation = 100 /4mg/dl.
If we standardize x to Z, then Z = ( x - ) / (2/n), where n is the number of samples taken , here 4 days.
Z = (x - 125)/ (100/4)
Therefore, P(x > 140) = P((x- 125)/ 25> (140- 125) /25)
= P( Z > 0.6)
From normal distribution table, for the area under the normal curve,
for Z= 0.6 , the area from 0 to 0.6 is 0.2258.
To get the required probability, we have to subtract this area from 0.5.
Therefore, P(x > 140) = P(Z > 1.5)
= 0.5 - 0.2558
= 0.2742
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