Question

PLEASE ANSWER QUICK!!! 20 POINTS!!!!1

Suppose the length of voicemails (in
seconds) is normally distributed with a mean
of 40 seconds and standard deviation of 10
seconds. Find the probability that a given
voicemail is between 20 and 50 seconds.
10
20
30
40
P = [?]%
Hint: use the 68 - 95 - 99.7 rule.
50 60
70
Enter

Answer 1

P=9600000000

Explanation:

1 point is =0.150119

40x10=400x20=8000x50x400000x10=400000x20=800000x30=240000000x40=9600000000

Answer 2

To find the probability that a given voicemail is between 20 and 50 seconds, we need to find the area under the normal curve between these two values.

Using the 68-95-99.7 rule, we know that approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

In this case, we want to find the probability that a voicemail falls within one standard deviation below the mean and one standard deviation above the mean, which corresponds to the interval (30,50).

To find this probability, we can use a standard normal distribution table, or we can convert the values to z-scores and use the standard normal distribution formula.

Using the formula, we have:

z1 = (20 - 40) / 10 = -2
z2 = (50 - 40) / 10 = 1

Now we look up the probability of a standard normal distribution between -2 and 1, which is approximately 0.818.

Therefore, the probability that a given voicemail is between 20 and 50 seconds is 81.8%.

P = 81.8%.