MATH SOLVE

2 months ago

Q:
# The lengths of songs on the radio are normally distributed with a mean length of 210 seconds. If 38.2% of all songs have lengths between 194 and 226 seconds, then the standard deviation of this distribution is:A) 16 seconds B) 8 seconds C) 32 seconds D) 64 seconds I’m suppose to construct a bell diagram for this question but don’t know how to without the standard deviation..... thx

Accepted Solution

A:

Given that the lengths of songs on the radio are normally distributed with a mean length of 210 seconds, then the probability that a randomly selected song will have length between 194 and 226 seconds is given by:

[tex]P(194\leq X\leq226)=P\left( \frac{194-210}{\sigma} \ \textless \ z\ \textless \ \frac{226-210}{\sigma} \right) \\ \\ =P\left(- \frac{16}{\sigma} \ \textless \ z\ \textless \ \frac{16}{\sigma} \right)=2P\left(z\ \textless \ \frac{16}{\sigma} \right)-1[/tex]

Given that the probability that all songs have lengths between 194 and 226 seconds is 38.2% = 0.382, then

[tex]2P\left(z\ \textless \ \frac{16}{\sigma} \right)-1=0.382 \\ \\ \Rightarrow2P\left(z\ \textless \ \frac{16}{\sigma} \right)=1+0.382=1.382 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{16}{\sigma} \right)= \frac{1.382}{2} =0.691 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{16}{\sigma} \right)=P(z\ \textless \ 0.5) \\ \\ \Rightarrow \frac{16}{\sigma} =0.5 \\ \\ \Rightarrow\sigma= \frac{16}{0.5} =32[/tex]

Therefore, the standard deviation of the distribution is 32 seconds.

[tex]P(194\leq X\leq226)=P\left( \frac{194-210}{\sigma} \ \textless \ z\ \textless \ \frac{226-210}{\sigma} \right) \\ \\ =P\left(- \frac{16}{\sigma} \ \textless \ z\ \textless \ \frac{16}{\sigma} \right)=2P\left(z\ \textless \ \frac{16}{\sigma} \right)-1[/tex]

Given that the probability that all songs have lengths between 194 and 226 seconds is 38.2% = 0.382, then

[tex]2P\left(z\ \textless \ \frac{16}{\sigma} \right)-1=0.382 \\ \\ \Rightarrow2P\left(z\ \textless \ \frac{16}{\sigma} \right)=1+0.382=1.382 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{16}{\sigma} \right)= \frac{1.382}{2} =0.691 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{16}{\sigma} \right)=P(z\ \textless \ 0.5) \\ \\ \Rightarrow \frac{16}{\sigma} =0.5 \\ \\ \Rightarrow\sigma= \frac{16}{0.5} =32[/tex]

Therefore, the standard deviation of the distribution is 32 seconds.