Question

b) the area to the gut UI c) the area that lies between - 1.02 and 0.89. Dc -0.73 to 2 1.35, d) the area that lies outside the interval from 2 nominal Cdbl-.13,l. ,0,1) 2. Suppose that the heights of pecan trees in a specific region are normally distributed with a mean of 41 feet and a standard deviation of 8.2 feet. a) What is the 90th percentile of the heights of the pecan trees? Round to one decimal place and include correct units. b) What is the probability that a randomly selected pecan tree has a height between 39 feet and 42 feet. Round to four decimal places. c) What is the probability that a randomly selected pecan tree has a height greater than 45 feet? Round to four decimal places. d) Would it be unusual for a randomly selected pecan tree to have a height greater than 45 feet? (Use the textbook's definition of unusual.) e) What is the probability that the mean height of a simple random sample of 36 pecan trees will be greater than 45 feet? Round to four decimal places. f) Would it be unusual for the mean height of a simple random sample of 36 pecan trees to be greater than 45 feet? (Use the textbook's definition of unusual.)

Answer 1

Mean = $\mu$ = 41

Standard deviation = $\sigma$ = 8.2

a)

We have given P(X < x) = 0.90

z value 0.90 is 1.28

We have to find the value of x

I= + 2*0= 41 + 1.28 * 8.2 = 51.5

b)

We have to find P(39 < X < 42)

For finding this probability we have to find a z score.

α –μ 39 - 41 8.2 = -0.24

1 –μ 42 – 41 820.12

That is we have to find P( - 0.24 < Z < 0.12)

P( - 0.24 < Z < 0.12) = P(Z < 0.12) - P(Z < - 0.24) = 0.5485 - 0.4037 = 0.1449

( From z table)

c)

We have to find P(X > 45)

For finding this probability we have to find z score.

α – μ 2 =- 45 - 41 - = 0.49 8.2

That is we have to find P(Z > 0.49)

P(Z > 0.49) = 1 - P(Z < 0.49) = 1 - 0.6872 = 0.3128

( From z table)

d)

If the probability that pecan tree to have a height greater than 45 feet is less than 0.05 we say that it is unusual.

But P(X > 45) = 0.3128 > 0.05 so it is not unusual.

e)

Sample size = n = 36

We have to find P( $\bar{x}$ > 45)

For finding this probability we have to find z score.

urlo

45-41 18. 236 1.3667

That is we have to find P(Z > 2.93)

P(Z > 2.93) = 1 - P(Z < 2.93) = 1 - 0.9983 = 0.0017

( From z table)

f)

If the probability that mean height pecan tree 45 feet is less than 0.05 we say that it is unusual.

In this case P($\bar{x}$ > 45) = 0.0.0017 < 0.05 so it is unusual.