Question

Q4. The haemoglobin levels for healthy adult females are approximately normally distributed with a mean of ? = 17 and standard deviation ? = 0.9. X ~ N(17, 0.9^2) a. Find the probability that a randomly chosen healthy female adult has a haemoglobin level i. Less than 15 ii. Between 16 and 18 iii. More than 17.5 b. Find the probability that the mean haemoglobin level of a group of five adult females will be above 17.5 c. Find the probability that the mean haemoglobin of a group of 20 female adults will be between 17.5 and 17.75 d. Find the haemoglobin level X for which 90% of the adult females have haemoglobin level less than X

          Q4. The haemoglobin levels for healthy adult females are approximately normally distributed with a mean of ? = 17 and standard deviation ? = 0.9.
X ~ N(17, 0.9^2)
a. Find the probability that a randomly chosen healthy female adult has a haemoglobin level
i. Less than 15
ii. Between 16 and 18
iii. More than 17.5
b. Find the probability that the mean haemoglobin level of a group of five adult females will be above 17.5
c. Find the probability that the mean haemoglobin of a group of 20 female adults will be between 17.5 and 17.75
d. Find the haemoglobin level X for which 90% of the adult females have haemoglobin level less than X

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Q4. The haemoglobin levels for healthy adult females are approximately normally distributed with a mean of ? = 17 and standard deviation ? = 0.9.
X N(17, 0.9^2)
a. Find the probability that a randomly chosen healthy female adult has a haemoglobin level
i. Less than 15
ii. Between 16 and 18
iii. More than 17.5
b. Find the probability that the mean haemoglobin level of a group of five adult females will be above 17.5
c. Find the probability that the mean haemoglobin of a group of 20 female adults will be between 17.5 and 17.75
d. Find the haemoglobin level X for which 90% of the adult females have haemoglobin level less than X

Added by Mary A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Supreeta N

05/16/2023

Step 1

z = (X - μ) / σ z = (15 - 17) / 0.9 z = -2.22 From the standard normal distribution table, the probability for z = -2.22 is approximately 0.0132. So, the probability that a randomly chosen healthy female adult has a haemoglobin level less than 15 is 0.0132. b. Show more…

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The haemoglobin levels for healthy adult females are approximately normally distributed with a mean of μ = 17 and standard deviation σ = 0.9. X ~ N(17, 0.9^2). Find the probability that a randomly chosen healthy female adult has a haemoglobin level: a) Less than 15 b) Between 16 and 18 c) More than 17.5 Find the probability that the mean haemoglobin level of a group of five adult females will be above 17.5. Find the probability that the mean haemoglobin of a group of 20 female adults will be between 17.5 and 17.75. Find the haemoglobin level X for which 90% of the adult females have haemoglobin level less than X.

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Transcript

00:01 Hi, today we are solving the question in which first of all we need to find the probability of x is smaller than 15.

00:09 So it is given by 5 into 15 plus 7 by 0 .9.

00:21 On solving it we get it equals to 0 .0131.

00:25 So therefore we get probability of x is small than 15 is equals to 0 .0131.

00:34 Now similarly secondly we need to find probability when x is lying between 16 and 18.

00:42 So if we do so, so 5 into 16 18 minus 17 divided by 0 .9 minus 5 into 17 16 minus 17 divided by 0 .9 on solving it we get 0 .8667 minus 0 .133.

01:07 So it is equals to 0 .7334.

01:12 From here we get therefore required probability when x is lying between 16 and 18 is equals to 0 .7334.

01:23 Now the next one is we need to find the probability of x is greater than 17 .5.

01:34 So to do so 1 minus 5 into 17 .5 minus 17 by 0 .19 0 .9 not 190 .9.

01:46 So it will be equal to 1 minus 0 .7107.

01:51 So it is equals to 0 .2893.

01:54 Now from here we get the required probability as therefore probability of x is greater than 17 .5 is equals to 0 .2893.

02:07 Now similarly we need to find the probability when x is greater than 17 .5.

02:25 So here 1 minus 5 into 17 .5 minus 17 divided by 0 .9 by root 5.

02:39 So here it is equals to 1 minus 0 .8929.

02:50 Here we can take it as 5z.

02:54 So we can see from here 0 .1071...

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