Question

Let x? be the mean of a random sample of size 50 drawn from a population with mean 112 and standard deviation 40. a. Find the mean and standard deviation of x?. b. Find the probability that x? assumes a value between 110 and 114. c. Find the probability that x? assumes a value greater than 113. Solution a. By the formulas in the previous section ?_x? = ? = 112 and ?_x? = ? / ? n = 40 / ? 50 = 5.65685 b. Since the sample size is at least 30, the Central Limit Theorem applies: x? is approximately normally distributed. We compute probabilities using Figure 12.2 "Cumulative Normal Probability" in the usual way, just being careful to use ?_x? and not ? when we standardize: P(110 < x? < 114) = P((110-?_x?)/?_x? < Z < (114-?_x?)/?_x?) = P((110-112)/5.65685 < Z < (114-112)/5.65685) = P(-0.35 < Z < 0.35) = 0.6368 - 0.3632 = 0.2736 c. Similarly P(x? > 113) = P(Z > (113-?_x?)/?_x?) = P(Z > (113-112)/5.65685) = P(Z > 0.18) = 1 - P(Z < 0.18) = 1 - 0.5714 = 0.4286

          Let x? be the mean of a random sample of size 50 drawn from a population with mean 112 and standard deviation 40.
a. Find the mean and standard deviation of x?.
b. Find the probability that x? assumes a value between 110 and 114.
c. Find the probability that x? assumes a value greater than 113.

Solution
a. By the formulas in the previous section
?_x? = ? = 112 and ?_x? = ? / ? n = 40 / ? 50 = 5.65685
b. Since the sample size is at least 30, the Central Limit Theorem applies: x? is approximately normally distributed. We compute probabilities using Figure 12.2 "Cumulative Normal Probability" in the usual way, just being careful to use ?_x? and not ? when we standardize:
P(110 < x? < 114) = P((110-?_x?)/?_x? < Z < (114-?_x?)/?_x?)
= P((110-112)/5.65685 < Z < (114-112)/5.65685)
= P(-0.35 < Z < 0.35) = 0.6368 - 0.3632 = 0.2736
c. Similarly
P(x? > 113) = P(Z > (113-?_x?)/?_x?)
= P(Z > (113-112)/5.65685)
= P(Z > 0.18)
= 1 - P(Z < 0.18) = 1 - 0.5714 = 0.4286

Show more…

Let x? be the mean of a random sample of size 50 drawn from a population with mean 112 and standard deviation 40.
a. Find the mean and standard deviation of x?.
b. Find the probability that x? assumes a value between 110 and 114.
c. Find the probability that x? assumes a value greater than 113.

Solution
a. By the formulas in the previous section
?x? = ? = 112 and ?x? = ? / ? n = 40 / ? 50 = 5.65685
b. Since the sample size is at least 30, the Central Limit Theorem applies: x? is approximately normally distributed. We compute probabilities using Figure 12.2 "Cumulative Normal Probability" in the usual way, just being careful to use ?x? and not ? when we standardize:
P(110 < x? < 114) = P((110-?x?)/?x? < Z < (114-?x?)/?x?)
= P((110-112)/5.65685 < Z < (114-112)/5.65685)
= P(-0.35 < Z < 0.35) = 0.6368 - 0.3632 = 0.2736
c. Similarly
P(x? > 113) = P(Z > (113-?x?)/?x?)
= P(Z > (113-112)/5.65685)
= P(Z > 0.18)
= 1 - P(Z < 0.18) = 1 - 0.5714 = 0.4286

Added by Jennifer B.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Marc Lauzon

12/02/2021

Step 1

The mean (μ) of the sample is the same as the mean of the population, which is 112. Show more…

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Let x̄ be the mean of a random sample of size 50 drawn from a population with mean 112 and standard deviation 40. a. Find the mean and standard deviation of x̄. b. Find the probability that x̄ assumes a value between 110 and 114. c. Find the probability that x̄ assumes a value greater than 113. Solution a. By the formulas in the previous section μ_x̄ = μ = 112 and σ_x̄ = σ / ∙n = 40 / √50 = 5.65685 b. Since the sample size is at least 30, the Central Limit Theorem applies: x̄ is approximately normally distributed. We compute probabilities using Figure 12.2 "Cumulative Normal Probability" in the usual way, just being careful to use σ_x̄ and not σ when we standardize: P(110 < x̄ < 114) = P((110-μ_x̄)/σ_x̄ < Z < (114-μ_x̄)/σ_x̄) = P((110-112)/5.65685 < Z < (114-112)/5.65685) = P(-0.35 < Z < 0.35) = 0.6368 - 0.3632 = 0.2736 c. Similarly P(x̄ > 113) = P(Z > (113-μ_x̄)/σ_x̄) = P(Z > (113-112)/5.65685) = P(Z > 0.18) = 1 - P(Z < 0.18) = 1 - 0.5714 = 0.4286

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Transcript

00:02 To answer these questions, we're going to be using the central limit theorem, which says, is we repeatedly take samples of size 50, the averages for all those little samples would form a bell curve with the center right around the population mean 112, and the standard deviation of all those little samples is going to equal the population standard deviation, which is 40, divided by the square root of 50.

00:30 And when we calculate that, we get a standard deviation of 5 .65685 -4 -249.

00:42 So i'm going to come and number my bell curve by 5 .66 to the right and to the left.

00:56 And now we want to figure out the probability that our sample that we pick, the average comes out to be between 110 and 114.

01:06 So 110 would fall right about here and 114 would fall right about here.

01:17 And we want the area in between these two markers.

01:21 So we're going to need a z score for 110.

01:26 So we'll do 110 minus the mean divided by that standard deviation number.

01:33 And that's going to give us a z score of negative 0 .35.

01:44 And then i'll look at the standard normal probability table, find a z score of negative 0 .35, and the area to the left of a z score of negative 0 .35 is 0 .3632.

01:59 So the area going this way is 0 .3632.

02:03 Now i also need a z score for 114.

02:07 So it's going to be 114 minus 112 divided by that standard deviation, which gives us a z score for 114...

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