Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 65 inches and standard deviation 4 inches. (a) What is the probability that an 18-year-old man selected at random is between 64 and 66 inches tall? (Round your answer to four decimal places.) (b) If a random sample of seven 18-year-old men is selected, what is the probability that the mean height x is between 64 and 66 inches? (Round your answer to four decimal places.) (c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this? The probability in part (b) is much higher because the mean is larger for the x distribution. The probability in part (b) is much higher because the standard deviation is larger for the x distribution. The probability in part (b) is much higher because the mean is smaller for the x distribution. The probability in part (b) is much lower because the standard deviation is smaller for the x distribution. The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.

Answer 1

Answer:

Step-by-step explanation:

a)

• Given that ; X ~ N ( µ = 65 , σ = 4 )

• P ( 64 < X < 66 )

From application of normal distribution ;

• Z = ( X - µ ) / σ, Z = ( 64 - 65 ) / 4,  Z = -0.25

• Z = ( 66 - 65 ) / 4,  Z = 0.25

Hence, P ( -0.25 < Z < 0.25 ) = P ( 64 < X < 66 ) = P ( Z < 0.25 ) - P ( Z < -0.25 ) P ( 64 < X < 66 ) = 0.5987 - 0.4013

• P ( 64 < X < 66 ) = 0.1974

b) X ~ N ( µ = 65 , σ = 4 )

• P ( 64 < X < 66 )

From normal distribution application ;

• Z = ( X - µ ) / ( σ / √(n)), plugging in the values,

• Z = ( 64 - 65 ) / ( 4 / √(12)) = Z = -0.866

• Z = ( 66 - 65 ) / ( 4 / √(12)) = Z = 0.866

P ( -0.87 < Z < 0.87 )

• P ( 64 < X < 66 ) = P ( Z < 0.87 ) - P ( Z < -0.87 )

• P ( 64 < X < 66 ) = 0.8068 - 0.1932

• P ( 64 < X < 66 ) = 0.6135

c) From the values gotten for (a) and (b), it is indicative that the probability in part (b) is much higher because the standard deviation is smaller for the x distribution.