Question

a. A normally distributed population has a mean of 98.36 and a standard deviation of 0.38. Determine the sample mean at the 80th percentile mark for samples of size 68.

Round to the nearest hundredth

b.

A normally distributed population has a mean of 98.15 and a standard deviation of 0.5. Determine the sample mean at the 20th percentile mark for samples of size 65.

Round to the nearest hundredth

c. A normally distributed population has a mean of 121 and a standard deviation of 22. Determine the value of the sample average at the 25th percentile for samples of siz 120.

Round to the nearest tenth

Answer 1

Solution :

a.

= / n = 0.38 / 68 = 0.0461

Using standard normal table ,  

P(Z < z) = 80%

P(Z < 0.84) = 0.8

z = 0.84

Using z-score formula,  

= z * + = 0.84 * 0.0461 + 98.36 = 98.4

80th percentile = 98.4

b.

= / n = 0.5 / 65 = 0.0620

Using standard normal table ,  

P(Z < z) = 20%

P(Z < -0.84) = 0.2

z = -0.84

Using z-score formula,  

= z * + = -0.84 * 0.0620 + 98.15 = 98.1

20th percentile = 98.1

c.

= / n = 22 / 120 = 2.0083

Using standard normal table ,  

P(Z < z) = 25%

P(Z < -0.67) = 0.25

z = -0.67

Using z-score formula,  

= z * + = -0.67 * 2.0083 + 121 = 119.7

25th percentile = 119.7