Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 125 millimeters, and a standard deviation of 7 millimeters.
If a random sample of 44 steel bolts is selected, what is the probability that the sample mean would be greater than 127.1 millimeters? Round your
answer to four
decimal places

P(X > 127.1) = 0.0233

Step-by-step explanation:

We are given;

Population mean; μ = 125 mm

Sample mean; x¯ = 127.1 mm

Sample size; n = 44

Standard deviation; σ = 7 mm

Formula for test statistic is;

z = (x¯ - μ)/(σ/√n)

z = (127.1 - 125)/(7/√44)

z = 2.1/1.0553

z = 1.99

From z-table attached;

P(X > 127.1) = P(z > 1.99) = 1 - 0.97670 = 0.0233

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation:

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 ☺