The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.6 cm a. Find the probability that an individual distance is greater than 214.80 cm.
b. Find the probability that the mean for 1515 randomly selected distances is greater than 204.00 cm
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
a. The probability is
(Round to four decimal places as needed.)
b. The probability is
(Round to four decimal places as needed.)
c. Choose the correct answer below.
A. The normal distribution can be used because the original population has a normal distribution.
B. The normal distribution can be used because the probability is less than 0.5
C. The normal distribution can be used because the mean is large.
D. The normal distribution can be used because the finite population correction factor is small

(a) the probability that an individual distance is greater than 214.80 cm is 0.1401.

(b) The probability that the mean for 15 randomly selected distances is greater than 204.00 cm is 0.2482.

(c) The normal distribution can be used because the original population has a normal distribution.

Step-by-step explanation:

We are given that the overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.6 cm.

(a) Let X = the overhead reach distances of adult females.

The z-score probability distribution for the normal distribution is given by;

                         Z  =    ~ N(0,1)  

where, = population mean reach distance = 205.5 cm  

            = standard deviation = 8.6 cm

So, X ~ Normal()

Now, the probability that an individual distance is greater than 214.80 cm is given by = P(X > 214.80 cm)

    P(X > 214.80 cm) = P( > ) = P(Z > 1.08) = 1 - P(Z 1.08)

                                                                    = 1 - 0.8599 = 0.1401

The above probability is calculated by looking at the value of x = 1.08 in the z table which has an area of 0.8599.

(b) Let = the sample mean selected distances.

The z-score probability distribution for the sample mean is given by;

                                   Z  =    ~ N(0,1)  

where, = population mean reach distance = 205.5 cm  

            = standard deviation = 8.6 cm

           n = sample size = 15

Now, the probability that the mean for 15 randomly selected distances is greater than 204.00 cm is given by = P( > 204.00 cm)

    P( > 204 cm) = P( > ) = P(Z > -0.68) = 1 - P(Z 0.68)

                                                                    = 1 - 0.7518 = 0.2482

The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.7518.

(c) The normal distribution can be used in part (b), even though the sample size does not exceed 30 because the original population has a normal distribution and the sample of 15 randomly selected distances has been taken from the population itself.

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺