**Please answer both questions** Question #1: four parts Question #2: one part I

Question

**Please answer both questions**
Question #1: four parts
Question #2: one part

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. It is estimated that 3.4% of the general population will live past their 90th birthday. In a graduating class of 757 high school seniors, find the following probabilities. (answers to four decimal places.) a) 15 or more will live beyond their 90th birthday b) 30 or more will live beyond their 90th birthday c) between 25 and 35 will live beyond their 90th birthday d) more than 40 will live beyond their 90th birthday Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean mu = 7050 and estimated standard deviation alpha = 2650. A test result of x

Explanation / Answer

Using Continuity Correction

E(X) = n p = 757 *0.034 = 25.738

sd(X) = sqrt(npq) = sqrt(757 *0.034 *0.966) = 4.98627195408

P(X>= 15 ) = P(X> 14.5) = P(Z>(14.5 - 25.738)/4.98627195408

= P(Z> -2.25378)= 0.9879

b) P(X>= 30)

P(X> 29.5) = P(Z>(29.5 - 25.738)/4.98627195408

= P(Z> 0.75447148383)= 0.2253

c)

P(25 <X< 35)

= P( 24.5 - 25.738)/4.98627195408 < Z< (35.5 - 25.738)/4.98627195408)

=P( -0.24828168 < Z< 1.9577752858)

=0.5729

d)

P(X>=40)

= P(X> 39.5) = P(Z>(39.5 - 25.738)/4.98627195408

= P(Z> 2.75997782045)= 0.0029

2)

Z = (X - 7050)/2650

P(X< 3500)

=P(Z< (3500 - 7050)/2650)

=P(Z< -1.339622)

=0.09019

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