overhead reach distances of adult females are normally distributed with a mean o

Question

overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.6 cm a. Find the probability that an individual distance is greater than 217.50 cm C. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? a.The probability is b. Find the probability that the mean for 25 randomly selected distances is greater than 203.70 cm Round to four decimal places as needed.) Enter your answer in the answer box and then click Check Answer 2 remaining

Explanation / Answer

a) P(X > 217.5)

= P((X - mean)/sd > (217.5 - mean)/sd)

= P(Z > (217.5 - 205)/8.6)

= P(Z > 1.45)

= 1 - P(Z < 1.45)

= 1 - 0.9265

= 0.0735

b) P(X > 203.7)

= P((X - mean)/(sd/sqrt(n)) > (203.7 - mean)/(sd/sqrt(n)))

= P(Z > (203.7- 205)/(8.6/sqrt(25)))

= P(Z > -0.76)

= 1 - P(Z < -0.76)

= 1 - 0.2236

= 0.7764

c) Because the main population distribution is normally distributred.

2) a) P(X > 308)

= P((X - mean)/sd > (308 - mean)/sd)

= P(Z > (308 - 266)/15)

= P(Z > 2.8)

= 1 - P(Z < 2.8)

= 1 - 0.9974

= 0.0026

b) P(X < x) = 0.04

or, P((X - mean)/sd < (x - 266)/15) = 0.04

or, P(Z < (x - 266)/15) = 0.04

or, (x - 266)/15 = -1.75

or, x = -1.75 * 15 + 266

or, x = 239.75

3)a) P(X < 100.6)

= P((X - mean)/sd < (100.6 - 98.19)/0.63)

= P(Z < 3.83)

= 1 = 100%

As the percentage is greater than 95%, so it is not appropriate.

b) P(X > x) = 0.05

or, P((X - mean)/sd > (x - 98.19)/0.63) = 0.05

or, P(Z > (x - 98.19)/0.63) = 0.05

or, P(Z < (x - 98.19)/0.63) = 0.95

or, (x - 98.19)/0.63) = 1.645

or, x = 1.645 * 0.63 + 98.19

or, x = 99.226

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