5. It is known that the certain pull-strength measurement is normally a. What is

Question

5. It is known that the certain pull-strength measurement is normally a. What is the probability that a randomly selected measurement is b. What is the probability that the sample mean found from a random distributed with mean = 10 lb, and the standard deviation = 1.2 lb greater than 10. 525 lb.? sample of size 10 is greater than 10. 525 lb.? C. If the population standard deviation is unknown, and s = 1.2 lb. is the standard deviation found from the sample, what is the probability that the sample mean found from a random sample of size 10 is greater than 10. 525 lb.? d. If the population standard deviation is unknown, and s = 1.2 lb, is the standard deviation found from the sample, what is the probability that the sample mean found from a random sample of size 50 is greater than 10. 525 lb.?

Explanation / Answer

SolutionA:

WILL SOLVE THIS PROBLEM USING TI83 CALC AND MANUAL STEP BY STEP.BOTH ANSWERS ARE SAME.

used TI83 calc

Step 1: Press the 2nd key and then press VARS then 2 to get “normalcdf.”

Step 2: Enter the following numbers into the screen:
10.525for the lower bound, followed by a comma, then 99999for the upper bound, followed by another comma.

Step 3: Press 10(for the mean), followed by a comma and then 1.2(for the standard deviation).

Step 4: Close the argument list with a “)”. (Your display should now read normalcdf(10.525, 99999, 10, 1.2).)

ANSWER:

0.3309

Solutionb:

std error=1.2/sqrt(10)

=0.379473

normalcdf(10.525, 99999, 10, 0.379473.)

=0.0833

ANSWER:0.0833

Solutionc:

std error=s/sqrt(samplesize

=1.2/sqrt(10)

=0.379473

normalcdf(10.525, 99999, 10, 0.379473.)

ANSWER:0.0833

Solutiond:

s=1.2

stderror=1.2/sqrt(50)

=0.169706

normalcdf(10.525, 99999, 10, 0.169706.)

=9.888*10^-4

=0.0009888

=0.001

Stepwise explanation without calc:

mean=10

stddev=1.2

P(X>10.525)

convert to z

z=x-mean/sd

=10.525-10/1.2

=0.4375

P(Z>0.4375)

=1-P(Z<0.4375)

=1-0.67

=0.33

Similarly

SOlutionb:

P(x bar>10.525)

z=10.525-10/1.2/sqrt(10)

=1.38

P(Z>1.38)

=1-P(Z<1.38)

=1-0.9162

=0.0838

Similarly

SolutionC:

mean=10

s=1.2

P(x bar>10.525)

z=10.525-10/1.2/sqrt(10)

=1.38

P(Z>1.38)

=1-P(Z<1.38)

=1-0.9162

=0.0838

SOlutiond:

mean=10

sd=1.2

n=50

Z=10.525-10/1.2/sqrt(50)

=3.094

P(Z>3.094)

=1-P(Z<3.094)

=1-0.999

=0.001

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.