At the end of the spring semester, the dean of students sent a survey to the entire freshman class. One question asked the students how much weight they had gained or lost since the beginning of the school year. The average was a gain of µ = 9 pounds with a standard deviation of LaTeX: \sigmaσ = 6. The distribution of scores was approximately normal. A sample of n = 4 students is selected, and the average weight change is computed for the sample. What is the probability that the sample mean will be greater than M = 10 pounds? In symbols, what is p(M > 10) (four decimals)

The probability of the average weight of the sample is greater than 10 pounds is 0.371.

Step-by-step explanation:

Hello!

The study variable is:

X: weight gained/lost since the beginning of the school year.

If this variable has a normal distribution with an average μ= 9 pounds and standard deviation δ= 6 pounds and a random sample of 4 freshmen was taken:

What is the probability that the sample mean will be greater than 10 pounds, symbolically:

P(X[bar]>10)

If the variable has a normal distribution X~N(μ;σ²) then the sample mean has also a normal distribution X[bar]~N(μ;σ²/n)

To calculate this probability you have to use the standard normal distribution.

Since the tables for this distribution show cumulative probabilities, you can rewrite the asked probability as:

P(X[bar]>10)= 1 - P(X[bar]≤10)

In other words, to know what is the probability "above" 10 pounds, you have to subtract the cumulated probability until 10 pounds to the total probability.

Next step is to standardize the value of the sample mean so that you can look for the probability using the distribution:

1 - P(X[bar]≤10)= 1 - P(Z≤(10-9)/(6/√4))= 1 - P(Z≤0.33)= 1 - 0.629= 0.371

The probability of the average weight of the sample is greater than 10 pounds is 0.371.

I hope it helps!

b

step-by-step explanation:

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632