The overhead reach distances of adult females are normally distributed with a mean of 205 cm205 cm and a standard deviation of 8.6 cm8.6 cm.
a. find the probability that an individual distance is greater than 215.00215.00 cm.
b. find the probability that the mean for 2525 randomly selected distances is greater than 203.70 cm.203.70 cm.
c. why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

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.

5 x 14?

im new, sorry if this is wrong.

step-by-step explanation:

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)