Suppose x has a distribution with a mean of 50 and a standard deviation of 4. Random samples of size n = 64 are drawn. (a) Describe the x distribution and compute the mean and standard deviation of the distribution. x has distribution with mean μx = and standard deviation σx = . (b) Find the z value corresponding to x = 51. z = (c) Find P(x < 51). (Round your answer to four decimal places.) P(x < 51) =

(a)

(b) The z-score for the sample mean = 51 is 2.

(c) The value of is 0.9773.

Step-by-step explanation:

The random variable X has mean, μ = 50 and standard deviation, σ = 4.

A random sample of size n = 64 is selected.

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the  distribution of sample mean is given by,

And the standard deviation of the  distribution of sample mean is given by,

The sample of X selected is, n = 64 > 30.

So, the Central limit theorem can be applied to approximate the distribution of sample mean ().

(b)

The z-score for the sample mean is given as follows:

Compute the z-score for = 51 as follows:

Thus, the z-score for the sample mean = 51 is 2.

(c)

Compute the value of as follows:

*Use a z-table for the probability.

Thus, the value of is 0.9773.

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)