2. P337. 23. A large box contains 10,000 ball bearings. A random sample of 120 is chosen. The sample mean diameter is 10 mm, and the standard deviation is 0.24 mm. True or false: (a) A 95% con dence interval for the mean diameter of the 120 bearings in the sample is 1 (1:96)(0:24)= p 120. (b) A 95% con dence interval for the mean diameter of the 10,000 bearings in the sample is 1 (1:96)(0:24)= p 120. (c) A 95% con dence interval for the mean diameter of the 10,000 bearings in the sample is 1 (1:96)(0:24)= p 10; 000.

B is True

A, C. D are false

Step-by-step explanation:

Given :

Sample size, n = 120

Mean diameter, m = 10

Standard deviation, s = 0.24

Confidence level, Zcritical ; Z0.05/2 = Z0.025 = 1.96

The confidence interval represents how the true mean value compares to a set of values around the mean computed from a set of sample drawn from the population.

The population here is N = 10000

To obtain

Confidence interval (C. I) :

Mean ± margin of error

Margin of Error = Zcritical * s/sqrt(n)

Margin of Error = 1.96 * 0.24/sqrt(120)

Confidence interval for the 10,000 ball bearing :

10 ± 1.96 * (0.24) / sqrt(120)

Hence. The confidence interval defined as :

10 ± 1.96 * (0.24) / sqrt(120) is the 95% confidence interval for the mean diameter of the 10,000 bearings in the box.

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation:

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.