Question

The height of 2-year old seedlings in a plant nursery are normally distributed with a population mean height of 11.0 cm. The population variance in height is 1.6 cm2. In a quality control check on the nursery a simple random sample of 15 seedlings was measured for height.

a) What is the expected value for the mean height of 15 sampled seedlings in this plant nursery?

b) What is the standard error for the mean height of 15 sampled seedlings in this plant nursery?

c) Sketch the distribution of the sample mean seedling height for 15 sampled seedlings. Make sure to label the x-axis with the population mean, the population mean +/- 1 standard deviation, the population mean +/- 2 standard deviations.

d) What is the probability of sampling an individual seedling with a height greater than 11.5 cm?

e) What is the probability of a sample of size 15 in this nursery having a mean height greater than 11.5 cm?

f) If you take 1000 samples of size 15 from this nursery, how many of those samples would you expect to have a mean height greater than 11.5 cm?

Answer 1

a)

Mean of samples = Mean of Population = 11 cm

b)

Std Dev of Population = 1.6 ½ = 1.265

Std Error of Sample = Std Dev of Population / Sample Size^1/2 = 1.265 / 15^1/2 = 0.327

c)

Mean +/- 1 SD = 11 +/- 0.327 = {10.67,11.33}

Mean +/- 2 SD = 11 +/- 2*0.327 = {10.35,11.65}

d)

P(X>11.5) = 1 – P(X<11.5) = 1 – P{z<(11.5-11)/1.265}

= 1 – P{z<0.40}   using the z-table

= 1 – 0.6554 = 0.3446 or 34.46%

e)

P(X>11.5) = 1 – P(X<11.5) = 1 – P{z<(11.5-11)/0.327}

= 1 – P{z<1.53}   using the z-table

= 1 – 0.937 = 0.063 or 6.3%

f)

Same as part e ie 6.3%