How do i go about solving this? i was sick and missed class, but the notes dont seem to be helping

Question

n a large population of adults, the mean IQ is 117 with a standard deviation of 15. Suppose 40 adults are randomly selected for a market research campaign. (Round all answers to 4 decimal places, if needed.)

(a) The distribution of IQ is (approximately norma,l is exactly normal, may or may not be normal ,is certainly skewed.)

(b) The distribution of the sample mean IQ is approximately normalexactly normalnot normalleft-skewedright-skewed with a mean of ? and a standard deviation of ?.

(c) The probability that the sample mean IQ is less than 116 is .?

(d) The probability that the sample mean IQ is greater than 116 is ?

(e) The probability that the sample mean IQ is between 116 and 119 is ?

how should i solve this?.

Answer 1

Hi Lyndsy,

(a) Tough call. In reality, no population is exactly normal, so true answer is approximately normal, but your book may define normality criteria as simply having known standard deviation and claim exactly normal. Real answer is approximately normal; not sure what the book wants.

(b) Approximately normal. Approximate normality requires large n and known population standard deviation. Now, there are differences in books in terms of what constitutes large n. For the purposes of this problem, we'll go with n>30. If your book goes with n>50, true answer is that we can't tell. As for the mean and standard deviation of the sample, again, if we assume 40 is sufficiently large, we can assume sample mean approximates population mean, so:

x-bar=mu=117

Now, standard deviation of a sample mean has its own formula and its own name, the standard error (SE):

SE=sigma/sqrt(n)

sigma=population standard deviation

n=sample size

For us:

sigma=15

n=40

SE=15/sqrt(40)

SE=2.3717

(c) Formula for computing these types of probabilities--comparing sample means against some given value--is:

z=(x-mu)/SE

Let's break this down.

z=z-score, which we can use to get the probability we were asked for

x=value we were given

mu=mean

SE=standard error

Now, for this particular problem:

x=116

mu=117

SE=2.3717

Thus:

z=(116-117)/2.3717

z= -0.42

Now, we go to the z-table, which can be found online or in any statistics textbook, go to column on the left for -0.4, go to row on top for 0.02. This gives 0.3372. This is called standardizing, which means in practice:

P(Z< -0.42)= P(X<116)=0.3372

(d) This is the complement rule or, as I call it, the "one minus trick." To compute P(X>a), when working with normal distributions, you need only take:

1-P(X<a)

Thus:

P(X>116)= 1-P(X<116)

P(X>116)=1-0.3372

P(X>116)=0.6628

(e) Similar to part (d), we will need to subtract here; however, it will be a different subtraction process. We need the probability that sample mean IQ is less than 119 first. Then, we can subtract the probability that it was less than 116, which we found above. Now, we go back to our formula:

z=(x-mu)/SE

x=119

mu=117

SE=2.3717

z=(119-117)/2.3717

z=0.84

Returning to our z-table and standardizing:

P(Z<0.84)=P(X<119)=0.7995

Now, we ultimately want:

P(116<X<119)

Note:

P(116<X<119) is equivalent to:

P( -0.42< Z < 0.84)

Do the subtraction:

P=0.7995-0.3372

P=0.4623

I hope this helps.