Respuesta :
Given the mean of 700 min. and the std. dev. of 120 min., we need to find the z-score for 940 minutes and then the area under the std. normal curve to the right of that z-score.
940 - 700
The applicable z score here is z = ----------------- = 2. What is the area under
120
the curve and to the right of 2 std. deviations from the mean?
You could find the answer from a table of z-scores, or you could take advantage of the empirical rule. The empirical rule states that 95% of the normally distributed data set lies within 2 std. dev. of the mean.
The area to the right of z=2 is 0.5 (half the area under the std. normal curve) less 0.95/2, that is, less 0.475.
Subtracting 0.475 from 0.500 yields 0.025 (answer). 2.5% of the time, the monthly phone bill will exceed 940, or 2 std. dev. above the mean.
940 - 700
The applicable z score here is z = ----------------- = 2. What is the area under
120
the curve and to the right of 2 std. deviations from the mean?
You could find the answer from a table of z-scores, or you could take advantage of the empirical rule. The empirical rule states that 95% of the normally distributed data set lies within 2 std. dev. of the mean.
The area to the right of z=2 is 0.5 (half the area under the std. normal curve) less 0.95/2, that is, less 0.475.
Subtracting 0.475 from 0.500 yields 0.025 (answer). 2.5% of the time, the monthly phone bill will exceed 940, or 2 std. dev. above the mean.
Answer:
the required probability = 0.25
Step-by-step explanation:
Mean = 700 minutes , Standard Deviation = 120 minutes
To find : the probability that more than 940 minutes were used.
Solution : First find the z-score for 940 minutes and then find the area under the curve to the right of associated standard deviations from the mean.
[tex]z-score=\frac{\mu_2-\mu_1}{\sigma}\\\\\mu_1=700,\mu_2=940,\sigma=120\\\\\implies z-score=\frac{940-700}{120}\\\\\implies z-score=2[/tex]
Now, calculate the area under the curve to the right of 2 standard deviations from the mean.
Take confidence level to be 95%.
The area to the right of 2 standard deviations is 0.5 that is half the area under the standard normal curve.
[tex]\implies Area < \frac{0.95}{2}\\\\\implies Area<0.475[/tex]
Now, to find the probability that more than 940 minutes were used = 0.5 - 0.475 = 0.25
Hence, the required probability = 0.25
