HELP ME PLEASE

7. When Mrs. Santo lived in China, temperature was measured in Celsius. She constantly had to convertir
Celsius to Fahrenheit until she started learning Celsius temperatures and how warm they are.
The formula to convert from Celsius to Fahrenheit is F = = C) + 32
The mean temperature for one month was 34°C and the standard deviation was 2.27.C. Applying what
know about multiplying by a constant and adding a constant, what was the mean and standard deviation
Fahrenheit? Be sure to show your work for both the mean and the standard deviation.

If a random variable [tex]Y[/tex] is obtained by linearly transforming [tex]X[/tex] for some constants [tex]m \ne 0[/tex] and [tex]b[/tex], then the mean of [tex]Y\![/tex] would be:

[tex]\overline{Y} = m\, \overline{X} + b[/tex].

[tex]\mathrm{Var}(Y) = m^{2} \, \mathrm{Var}(X)[/tex].

The standard deviation of a random variable is the square root of its variance. Therefore, the standard deviation of [tex]Y[/tex] would be:

[tex]\begin{aligned}\mathrm{SD}(Y) &= \sqrt{\mathrm{Var}(Y)} \\ &= \sqrt{m^{2}\, \mathrm{Var}(X)} \\ &= m\, \sqrt{\mathrm{Var}(X)} \\ &= m\, \mathrm{SD}(X)\end{aligned}[/tex].

In this question, the random variable [tex]F[/tex] is obtained by linearly transforming [tex]C[/tex] using the constants [tex]m = (9/5)[/tex] and [tex]b = 32[/tex]. That is:

[tex]F = (9/5)\, C + 32[/tex].

It is given that mean [tex]\overline{C} = 34[/tex] while standard deviation [tex]\mathrm{SD}(X) = 2.27[/tex]. The mean and standard deviation of [tex]F[/tex] obtained from the linear transformation would be:

[tex]\begin{aligned}\overline{F} &= m\, \overline{C} + b \\ &= (9/5) \times 34 + 32\\ &= 93.2\end{aligned}[/tex].

[tex]\begin{aligned}\mathrm{SD}(F) &= m\, \mathrm{SD}(C) \\ &= (9/5) \times 2.27 = 4.086\end{aligned}[/tex].

Thus, when measured in degrees Fahrenheit, the corresponding mean would be [tex]93.2\; \rm ^{\circ} F[/tex] and and standard deviation would be [tex]4.086\; \rm ^{\circ} F[/tex].