health care statistics problem?

What is the probability that the cost will be more than $500? (to 4 decimals)

z = (x - µ) /σ

z = (500 - 328) /92 = 1.869565

from z-score table

P( x ≤ 500) = 0.969228

P(x > 500) = 1 - P( x ≤ 500)

P(x > 500) = 0.0308

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What is the probability that the cost will be less than $250? (to 4 decimals)

z = (x - µ) /σ

z = (250 - 328) /92 = -0.8478261

z = (500 - 328) /92 = 1.869565

from z-score table

P( x < 250) = 0.1983

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What is the probability that the cost will be between $300 and $400? (in inches, to 4 decimals)

z = (300 - 328) /92 = -0.30435

P( x ≤ 300) = 0.380431

z = (400 - 328) /92 = 0.782609

P( x ≤ 400) = 0.783072

P( 300 < x < 400 )

= 0.783072 - 0.380431

= 0.402641

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If the cost to a patient is in the lower 8% of charges for this medical service, what was the cost of this patient’s emergency room visit? (to 2 decimals)

p = 8% = 0.08

then z = -1.40507

z = (x - µ) /σ

x = µ + σz

x = 328 + 92(-1.40507)

x = $198.73

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1

X follows a normal distribution

X~N(328,92)

Now probability that the cost will be more than $500

=P(x>500)

=1-P(X<=500)

=1-P((x-328)/92<=(500-328)/92)

=1-P(z<=1.8695)

=1-Ф(z)

=1-0.96925 (from z score table)

=0.0307

probability that the cost will be less than $250

=P(X<250)

=P(z<=(250-328)/92)

=P(z<= -0.847)

=Ф(-0.847)

=1-Ф(0.847)

=1-0.80233

=0.1976

probability that the cost will be between $300 and $400?

P(300<X<400)= P(X<400)-P(X>300)

=P(z<0.782)-P(z>-0.3043)

=P(z<0.782)-(1-P(z<= -0.3043))

=0.7823 – 1+Ф(-0.3043)

=0.7823-1+(1-Ф(0.3043))

=0.7823-Ф(0.3043)

=0.7823-0.6179

=0.1644

Here p=0.08

So, z =(x-μ)/σ =-1.405

Thus, x=μ+σz

=328+92*(-1.405)

=198.73

So,the cost is $198.73

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