Question

Assume the random variable X is normally distributed with mean μ= 50 and standard deviation σ 7. Find the 87th percentile. The 87th percentlie is Round to two decimal places as needed.)

Answer 1

1) P(X < x) = 0.87

or, P((X - $\mu$)/$\sigma$ < (x - $\mu$)/$\sigma$) = 0.87

or, P(Z < (x - 50)/7) = 0.87

or, (x - 50)/7 = 1.13

or, x = 1.13 * 7 + 50

or, x = 57.91

2)a) P(1100 < X < 1400)

= P((1100 - $\mu$)/$\sigma$< (X - $\mu$)/$\sigma$< (1400 - $\mu$)/$\sigma$)

= P((1100 - 1252)/129 < Z < (1400 - 1252)/129)

= P(-1.18 < Z < 1.15)

= P(Z < 1.15) - P(Z < -1.18)

= 0.8749 - 0.1190

= 0.7559

b) P(X < 1050)

= P((X - $\mu$)/$\sigma$ < (1050 - $\mu$)/$\sigma$)

= P(Z < (1050 - 1252)/129)

= P(Z < -1.57)

= 0.0582

c) P(X > 1200)

= P((X - $\mu$)/$\sigma$ > (1200 - $\mu$)/$\sigma$)

= P(Z > (1200 - 1252)/129)

= P(Z > -0.40)

= 1 - P(Z < -0.40)

= 1 - 0.3446

= 0.6554

d) P(X < 1050)

= P((X - $\mu$)/$\sigma$ < (1050 - $\mu$)/$\sigma$)

= P(Z < (1050 - 1252)/129)

= P(Z < -1.57)

= 0.0582 = 5.82%

3)a) P(X > 285)

= P((X - $\mu$)/$\sigma$ > (285 - $\mu$)/$\sigma$)

= P(Z > (285 - 279)/8)

= P(Z > 0.75)

= 1 - P(Z < 0.75)

= 1 - 0.7734

= 0.2266

b) P(267 < X < 283)

= P((267 - $\mu$)/$\sigma$ < (X - $\mu$)/$\sigma$ < (283 - $\mu$)/$\sigma$)

= P((267 - 279)/8 < Z < (283 - 279)/8)

= P(-1.5 < Z < 0.5)

= P(Z < 0.5) - P(Z < -1.5)

= 0.6915 - 0.0668

= 0.6247

c) P(X < 277)

= P((X - $\mu$)/$\sigma$< (277 - $\mu$)/$\sigma$)

= P(Z < (277 - 279)/8)

= P(Z < -0.25)

= 0.4013

d) P(X < 261)

= P((X - $\mu$)/$\sigma$ < (261 - $\mu$)/$\sigma$)

= P(Z < (261 - 279)/8)

= P(Z < -2.25)

= 0.0122

Since the probability is less than 0.05, so very preterm babies are unusual.