Secure https://www.mathxl.com/Student/PlayerHomework.aspx?homeworkld-464546945&q

Question

Secure https://www.mathxl.com/Student/PlayerHomework.aspx?homeworkld-464546945&questionid;=4&flushed;:false&cld-4834238;&centerwin;=yes ADMN 210, Winter 2018, Section 2, MW 4 PM Kathya N Homework: Homework Chapter 7 Score: 0.8 of 4 pts 1 of 6 (5 complete) &7.2.5-T The diameter of a brand of tennis balls is approximately normally distributed, with a mean of 2.53 inches and a standard deviation of 0.04 inch. A random sample of 11 tennis balls through (d) below a. What is the sampling distribution of the mean? OA. Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of size 11 will be the uniform distribution B. Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of size 11 will also be approximately normal. ° C. Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of size 11 cannot be found D. Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of size 11 will not be approximately normal b. What is the probability that the sample mean is less than 2.51 inches? P(X

Explanation / Answer

mean = 2.53
std. dev. = 0.04
n = 11

b)
As the sample size is less than 30, we use t-distribution to calculate the probability

Using central limit theorem,
for x = 2.51, t = (2.51 - 2.53)/(0.04/sqrt(11)) = -1.6583
df = 11 - 1 = 10

P(X<2.51)
= P(t < -1.6583)
= 0.0641 (using t-distribution table)

c)
As the sample size is less than 30, we use t-distribution to calculate the probability

Using central limit theorem,
for x = 2.52, t = (2.52 - 2.53)/(0.04/sqrt(11)) = -0.8292
for x = 2.55, t = (2.55 - 2.53)/(0.04/sqrt(11)) = 1.6583

df = 11 - 1 = 10

P(2.52 < X < 2.55)
= P(-0.8292 < t < 1.6583)
= P(t < 1.6583) - P(t < -0.8292)
= 0.9359 - 0.2132 (using t-distribution table)
= 0.7227

d)
63% = 0.63
0.63/2 = 0.315
For left tail the area which lies to the left of the value = 0.5 - 0.315 = 0.185
Using t-distribution table, for df = 10, value of t = -0.9387

Now using central limit theorem,
xbar = 2.53 - 0.9387*0.04 = 2.4925
xbar = 2.53 + 0.9387*0.04 = 2.5676

Lower bound = 2.4925 and upper bound = 2.5676

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---