Answer:
a) P(z>1.16) = 0.8770
b) P(z>-0.75) = 0.2266
Step-by-step explanation:
Mean = 205 cm
Standard Deviation = 8.6 cm
a) Find the probability that an individual distance is greater than 215.00
We need to find P(X>215)
x = 215
z = x - mean /standard deviation
z = 215 - 205 / 8.6
z = 1.16
P(X>215)=P(z>1.16)
Finding value of z =1.16 from the table
P(z>1.16) = 0.8770
b) Find the probability that the mean for 25 randomly selected distances is greater than 203.70 cm
Sample size n= 25
x = 203.70
mean = x- mean / standard deviation / √sample size
mean = 203.70 - 205 / 8.6 / √25
mean = -1.3/8.6/5
mean = -0.75
Finding value from z-score table
P(mean >-0.75) = 0.2266
c) Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
If the original problem is normally distributed, then for any sample size n, the sample means are normally distributed.
