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A doctor is researching the effects of "too much text messaging" on a group of teenagers. He has already shown that the average number of text messages sent per day by teenagers is 312 with a standard deviation of 125. He wants to know how many text messages separate the lowest 15% from the highest 85% in a sampling distribution of 144 teenagers. Use the [tex]$z$[/tex]-table below:

\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline
[tex]$z$[/tex] & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & 0.08 & 0.09 \\
\hline
-1.3 & 0.097 & 0.095 & 0.093 & 0.092 & 0.090 & 0.089 & 0.087 & 0.085 & 0.084 & 0.082 \\
\hline
-1.2 & 0.115 & 0.113 & 0.111 & 0.109 & 0.107 & 0.106 & 0.104 & 0.102 & 0.100 & 0.099 \\
\hline
-1.1 & 0.136 & 0.133 & 0.131 & 0.129 & 0.127 & 0.125 & 0.123 & 0.121 & 0.119 & 0.117 \\
\hline
-1.0 & 0.159 & 0.156 & 0.154 & 0.152 & 0.149 & 0.147 & 0.145 & 0.142 & 0.140 & 0.138 \\
\hline
-0.9 & 0.184 & 0.181 & 0.179 & 0.176 & 0.174 & 0.171 & 0.169 & 0.166 & 0.164 & 0.161 \\
\hline
-0.8 & 0.212 & 0.209 & 0.206 & 0.203 & 0.201 & 0.198 & 0.195 & 0.192 & 0.189 & 0.187 \\
\hline
-0.7 & 0.242 & 0.239 & 0.236 & 0.233 & 0.230 & 0.227 & 0.224 & 0.221 & 0.218 & 0.215 \\
\hline
\end{tabular}

Round the [tex]$z$[/tex]-score to two decimal places. Round [tex]$x$[/tex] to the nearest whole number.

Provide your answer below:

[tex]\[
\begin{array}{l}
z \text{-score} = \square \\
\bar{x} = \square \text{ text messages}
\end{array}
\][/tex]

Sagot :

GinnyAnswer
To determine how many text messages separate the lowest 15% from the highest 85% in a sampling distribution of 144 teenagers, we need to follow these steps carefully:

1. Identify the given parameters:
- Population mean (μ): 312
- Population standard deviation (σ): 125
- Sample size (n): 144
- Percentile to find the threshold for: 15%

2. Find the corresponding z-score for the 15th percentile using the z-table.
Looking up the z-table, we see that the z-score for the 15th percentile is approximately -1.04.

3. Calculate the standard deviation of the sampling distribution (standard error):
[tex]\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{125}{\sqrt{144}} = \frac{125}{12} \approx 10.42 \][/tex]

4. Calculate the value (x) that separates the lowest 15% using the z-score formula:
[tex]\[ x = \mu + z \cdot (\text{Standard Error}) \][/tex]
Replacing the known values we get:
[tex]\[ x = 312 + (-1.04) \cdot 10.42 \approx 312 - 10.83 \approx 301 \][/tex]

Therefore, the z-score and the corresponding number of text messages that separate the lowest 15% are:
[tex]\[ \begin{array}{l} z \text {-score }=-1.04 \\ \bar{x}=301 \text { text messages } \end{array} \][/tex]
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