Question

1.The height of an adult male in the United States is approximately normally distributed with a mean of 69.3 inches and a standard deviation of 2.8 inches. Find the percentile P76 for the heights of adult males in the United States.

Round Answer to 4 decimal places.

2. The height of an adult male in the United States is approximately normally distributed with a mean of 69.3 inches and a standard deviation of 2.8 inches. Assume that such an individual is selected at random. What is the probability that his height will be between 65.8 inches and 71.2 inches?

Round answer to 4 decimal places.

3.A certain standardized test has scores which range from 0 to 500, with decimal scores possible. Scores on the exam are normally distributed with a mean of 308 and a standard deviation of 45. What proportion of students taking the exam receive a score less than 361?

Round answer to 4 decimal places.

Answer 1

Solution:-

Given that,

mean = = 69.3 in.

standard deviation = = 2.8 in.

1) Using standard normal table,

P(Z < z) = 76%

= P(Z < z ) = 0.76

= P(Z < 0.71 ) = 0.76  

z = 0.71

Using z-score formula,

x = z * +

x = 0.71 * 2.8 + 69.3

x = 71.3 in.

P76 = 71.3 in.

2) P(65.8 < x < 71.2) = P[(65.8 - 69.3)/ 2.8) < (x - ) /  < (71.2 - 69.3) / 2.8 ) ]

= P(-1.25 < z < 0.68)

= P(z < 0.68) - P(z < -1.25)

Using z table,

= 0.7517 - 0.1056

= 0.6461

3) Given that ,

mean = = 308

standard deviation = = 45

P(x < 361)

= P[(x - ) / < (361 - 308) / 45]

= P(z < 1.18)

Using z table,

= 0.8810