A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly selected element X of the population is between 57,000 and 58,000. Find the mean and standard deviation of X - for samples of size 100. Find the probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; = 57,800  and   = 750.

Let X = randomly selected element of the population

The z probability is given by;

           Z = ~ N(0,1)  

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( <= ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( < ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = = = 75

The z probability is given by;

           Z = ~ N(0,1)  

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( <= ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( < ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

answer: yes, since the line has a steeper rate of change than the exponential the y-values will be greater and they will only intersect one time.

step-by-step explanation:

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