Exercise 30 Page 744

Let's look at the given data set to see if the classmate's claim is correct. There are two possible cases for the placement of 248 and 332. The first case is when 248 and 332 are on the different sides of the mean.

The other case is when both 248 and 332 are on the same side of the mean.

To show that the mean and standard deviation are, in fact, different in each case, we will calculate them. Let's start with the first case.

Case 1

Look at the graph of the normal curve, assuming that 248 and 332 are on the different sides of the mean. The value 248 is one standard deviation away from the mean, and 332 is three standard deviations away.

In this case we can write the following system of equations. mean-1 deviation=248 & (I) mean+3 deviations=332 & (II) For convenience, the mean will be represented by m and the standard deviation will be represented by s. m-s=248 & (I) m+3 s=332 & (II) Let's solve this system using substitution.

m-s=248 m+3 s=332

▼

(I): Solve for s

SubEqn

(I): LHS-m=RHS-m

- s=248-m m+3s=332

MultEqn

(I): LHS * (-1)=RHS* (-1)

s=-(248-m) m+3 s=332

Distr

(I): Distribute -1

s=-248+m m+3 s=332

We can now substitute s=-248+m into Equation (II).

s=-248+m m+3 s=332

Substitute

(II): s= -248+m

s=-248+m m+3 ( -248+m)=332

▼

(II): Solve for m

Distr

(II): Distribute 3

s=-248+m m-744+3m=332

AddEqn

(II): LHS+744=RHS+744

s=-248+m m+3m=332+744

AddTerms

(II): Add terms

s=-248+m 4m=1076

DivEqn

(II): .LHS /4.=.RHS /4.

s=-248+m m=269

We found that the mean m can equal to 269. Now to find the value of the standard deviation s, we can substitute m=269 into either of the equations in the given system. Let's use the first equation.

s=-248+m m=269

Substitute

(I): m= 269

s=-248+ 269 m=269

AddTerms

(I): Add terms

s=21 m=269

The mean is equal to 269 and the standard deviation is 21.

Case 2

This time, 248 and 332 are on the same side of the mean. The value 248 is one standard deviation away from the mean and 332 is three standard deviations away.

We can write the following system of equations. mean+1 deviation=248 & (I) mean+3 deviations=332 & (II) For convenience, the mean will be represented by m and the standard deviation will be represented by s. m+s=248 & (I) m+3 s=332 & (II) To solve this system, we can again use substitution.

m+s=248 m+3 s=332

SubEqn

(I): LHS-m=RHS-m

s=248-m m+3 s=332

We can now substitute s=248-m into Equation (II).

s=248-m m+3 s=332

Substitute

(II): s= s=248-m

s=248-m m+3 ( 248-m)=332

▼

(II): Solve for m

Distr

(II): Distribute 3

s=248-m m+744-3m=332

SubEqn

(II): LHS-744=RHS-744

s=248-m m-3m=332-744

AddSubTerms

(II): Add and subtract terms

s=248-m -2m=-412

DivEqn

(II): .LHS /-2.=.RHS /-2.

s=248-m m=206

This time, the mean m is equal to 206. Now to find the value of the standard deviation s, we can substitute m=206 into either of the equations in the given system. Let's use the first equation.

s=248-m m=206

Substitute

(I): m= 206

s=248- 206 m=206

SubTerm

(I): Subtract term

s=42 m=206

We found that in this case the mean is 206 and the standard deviatio is equal to 42.

Conclusion

We found two possible means and standard deviations for the given data set.

Solution 1		Solution 2
Mean 1	Standard deviation 1	Mean 2	Standard deviation 2
269	21	206	42

Therefore, the classmate is wrong.