Question

Find z such that 5.4% of the standard normal curve lies to the right of z. (Round your answer to two decimal places.)
z =  

Find the z value such that 83% of the standard normal curve lies between −z and z. (Round your answer to two decimal places.)
z =  

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.1 millimeters (mm) and a standard deviation of 0.7 mm. For a randomly found shard, find the following probabilities. (Round your answers to four decimal places.)

(a) the thickness is less than 3.0 mm


(b) the thickness is more than 7.0 mm


(c) the thickness is between 3.0 mm and 7.0 mm

Answer 1

Solution :

Given that,  

1.

Using standard normal table ,

P(Z > z) = 5.4%

1 - P(Z < z) = 0.054

P(Z < z) = 1 - 0.054

P(Z < 1.61) = 0.946

z = 1.61

2.

P(-z Z z) = 83%

P(Z z) - P(Z -z) = 0.83

2P(Z z) - 1 = 0.83

2P(Z z) = 1 + 0.83 = 1.83

P(Z z) = 1.83 / 2 = 0.915

P(Z 1.37) = 0.915

z = 1.37

3.

mean = = 5.1

standard deviation = = 0.7

(a)

P(x < 3.0) = P[(x - ) / < (3.0 - 5.1) / 0.7]

= P(z < -3)

= 0.0013

Probability = 0.0013

(b)

P(x > 7.0) = 1 - P(x < 7.0)

= 1 - P[(x - ) / < (7.0 - 5.1) / 0.7)

= 1 - P(z < 2.71)

= 1 - 0.9966

= 0.0034.

Probability = 0.0034

(c)

P(3.0 < x < 7.0) = P[(3.0 - 5.1)/ 0.7) < (x - ) /  < (7.0 - 5.1) / 0.7) ]

= P(-3 < z < 2.71)

= P(z < 2.71) - P(z < -3)

= 0.9966 - 0.0013

= 0.9953

Probability = 0.9953