First we calculate the z-score for this situation.

Finding <em>P</em>(x < 5.3) is the same as finding <em>P</em>(<em>z</em> < -0.8). Using a table of standard normal distribution and z-scores we see that the probability is 0.21186.
Answer:
≈0.487
Step-by-step explanation:
I think this is your full question right?
Jose surveyed the length of TV commercials (in seconds). Find, to 3 decimal places, the experimental probability that a randomly chosen TV commercial will last:
Length Frequency
0-19 17
20-39 38
40-59 19 20 to 39 seconds.
60+ 20 to 4
Here is my anwer:
experimental probability = relative frequency =
Total frequencies = 17 + 38 +18 + 4 = 78
P(20 to 39s) = 38/78 ≈0.487
Answer:
Step-by-step explanation:
B, y=5x
There are 730 total boxes. The last box has 108 toys.