Question

The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): Critical Reading 502 Mathematics 515 Writing 494Assume that the population standard deviation on each part of the test is = 100.a. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test (to 4 decimals)?b. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test (to 4 decimals)?c. What is the probability a sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test (to 4 decimals)?

Answer 1

a)
`P(492<X<512) = P((492-502)/(10.54) <Z<(512-502)/(10.54)) = P(-0.949 < Z< 0.949)`

And we can use excel or the normal standard table to find this probability:

`P(-0.949 < Z< 0.949)= P(Z<0.949) -P(Z<-0.949) =0.8287-0.1713=0.6574`

b)
`P(505<X<525) = P((505-515)/(10.54) <Z<(525-505)/(10.54)) = P(-0.949 < Z< 1.898)`

And we can use excel or the normal standard table to find this probability:

`P(-0.949 < Z< 1.898)= P(Z<1.898) -P(Z<-0.949) =0.9712-0.1713=0.7998`

c)
`P(484<X<504) = P((484-494)/(10) <Z<(504-494)/(10)) = P(-1 < Z< 1)`

And we can use excel or the normal standard table to find this probability:

`P(-1 < Z< 1)= P(Z<1) -P(Z<-1) =0.8413-0.1587=0.6827`

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the scores for critical reading of a population, and for this case we know the distribution for X is given by:

`X \sim N(502,100)`

Where
`\mu=502` and
`\sigma=100`

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

`\bar X \sim N(\mu, (\sigma)/(√(n))=(100)/(√(90))=10.54)`

And for this case we want this probability:

`P(502-10 < \bar X < 502+10)`

And for this case we can use the z score given by:

`z= (\bar X -\mu)/(\sigma_(\bar x))`

And if we use this formula we got:

`P(492<X<512) = P((492-502)/(10.54) <Z<(512-502)/(10.54)) = P(-0.949 < Z< 0.949)`

And we can use excel or the normal standard table to find this probability:

`P(-0.949 < Z< 0.949)= P(Z<0.949) -P(Z<-0.949) =0.8287-0.1713=0.6574`

Part b

Let X the random variable that represent the scores for Math of a population, and for this case we know the distribution for X is given by:

`X \sim N(515,100)`

Where
`\mu=515` and
`\sigma=100`

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

`\bar X \sim N(\mu, (\sigma)/(√(n))=(100)/(√(90))=10.54)`

And for this case we want this probability:

`P(515-10 < \bar X < 515+10)`

And for this case we can use the z score given by:

`z= (\bar X -\mu)/(\sigma_(\bar x))`

And if we use this formula we got:

`P(505<X<525) = P((505-515)/(10.54) <Z<(525-505)/(10.54)) = P(-0.949 < Z< 1.898)`

And we can use excel or the normal standard table to find this probability:

`P(-0.949 < Z< 1.898)= P(Z<1.898) -P(Z<-0.949) =0.9712-0.1713=0.7998`

Part c

Let X the random variable that represent the scores for Writing of a population, and for this case we know the distribution for X is given by:

`X \sim N(494,100)`

Where
`\mu=494` and
`\sigma=100`

We select a sample of size n=100, since the distribution for X is normal then the distribution for the sample size is also normal

`\bar X \sim N(\mu, (\sigma)/(√(n))=(100)/(√(100))=10)`

And for this case we want this probability:

`P(494-10 < \bar X < 494+10)`

And for this case we can use the z score given by:

`z= (\bar X -\mu)/(\sigma_(\bar x))`

And if we use this formula we got:

`P(484<X<504) = P((484-494)/(10) <Z<(504-494)/(10)) = P(-1 < Z< 1)`

And we can use excel or the normal standard table to find this probability:

`P(-1 < Z< 1)= P(Z<1) -P(Z<-1) =0.8413-0.1587=0.6827`