Now assume you take a sample of n = 7 lemons. a) What is the probability of the

Question

Now assume you take a sample of n = 7 lemons.

a) What is the probability of the sample average being within 10 grams of the population mean (100 g)? In other words, you need to figure out Pr{ - 10 < Y < + 10}

b) What is the probability of the sample average being within 7 grams of the population mean if the population mean is actually = 120g? Are you surprised? Why or why not? You should make sure you understand what happened here.

c) Now assume a sample of n = 21 lemons and repeat (a).

d) What is the effect of sample size?

Explanation / Answer

standard deviation is not given , so we are assuming the standard deviation to be 10 in this case

given n = 7

a) What is the probability of the sample average being within 10 grams of the population mean (100 g)

In other words, you need to figure out Pr{ - 10 < Y < + 10}

so P(90<X<110) , using th z score formula we get

Z = (X-Mean)/(SD/sqrt(n))

(90-100)/(10/sqrt(7)) <Z< (110-100)/(10/sqrt(7))

-2.64<Z<2.64 , please keep the z table handy now

To find the probability of P (2.64<Z<2.64), we use the following formula:

P (2.64<Z<2.64 )=P ( Z<2.64 )P (Z<2.64 )

P ( Z<2.64 ) can be found by using the following fomula.

P ( Z<a)=1P ( Z<a )

After substituting a=2.64 we have:

P ( Z<2.64)=1P ( Z<2.64 )

We see that P ( Z<2.64 )=0.9959 so,

P ( Z<2.64)=1P ( Z<2.64 )=10.9959=0.0041

At the end we have:

P (2.64<Z<2.64 )=0.9918

b)

performing the same calculation again but with different values

In other words, you need to figure out Pr{ - 7 < Y < + 7}

so P(93<X<107) , using th z score formula we get

Z = (X-Mean)/(SD/sqrt(n))

(93-120)/(10/sqrt(7)) <Z< (107-120)/(10/sqrt(7))

-7.14<Z<-3.43 , please keep the z table handy now

To find the probability of P (7.13<Z<3.43), we use the following formula:

P (7.13<Z<3.43 )=P ( Z<3.43 )P (Z<7.13 )

P ( Z<3.43 ) can be found by using the following fomula.

P ( Z<a)=1P ( Z<a )

After substituting a=3.43 we have:

P ( Z<3.43)=1P ( Z<3.43 )

We see that P ( Z<3.43 )=0.9997 so,

P ( Z<3.43)=1P ( Z<3.43 )=10.9997=0.0003

c)

changing n = 21 we get

n other words, you need to figure out Pr{ - 10 < Y < + 10}

so P(90<X<110) , using th z score formula we get

Z = (X-Mean)/(SD/sqrt(n))

(90-100)/(10/sqrt(21)) <Z< (110-100)/(10/sqrt(21))

-4.58 < Z< 4.58

To find the probability of P (4.58<Z<4.58), we use the following formula:

P (4.58<Z<4.58 )=P ( Z<4.58 )P (Z<4.58 )

P ( Z<4.58 ) can be found by using the following fomula.

P ( Z<a)=1P ( Z<a )

After substituting a=4.58 we have:

P ( Z<4.58)=1P ( Z<4.58 )

We see that P ( Z<4.58 )=1 so,

P ( Z<4.58)=1P ( Z<4.58 )=11=0

At the end we have:

P (4.58<Z<4.58 )=1

d) we see based on our calculations from a and c

the sample size increases , the probability also increases

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