Respuesta :
Answer:
The probability that the mean amplifier output would be greater than 449.8 watts in a sample of 76 amplifiers is 0.8078.
Step-by-step explanation:
According to the Central Limit Theorem if an unknown population is selected with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from this population with replacement, then the distribution of the sample means will be approximately normally. Â
Then, the mean of the sample means is given by,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the sample means is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
The information provided is as follows:
[tex]\mu=451\\\sigma^{2}=144\\n=76\\\bar x=449.8[/tex]
Compute the probability that the mean amplifier output would be greater than 449.8 watts in a sample of 76 amplifiers as follows:
[tex]P(\bar X>449.8)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{449.8-451}{\sqrt{144/76}})\\\\=P(Z>-0.87)\\\\=P(Z<0.87)\\\\=0.8078[/tex]
*Use a z-table.
Thus, the probability that the mean amplifier output would be greater than 449.8 watts in a sample of 76 amplifiers is 0.8078.
