Jamilla baby sat nine times. she earned $15, $20, $10, $12, $20, $16, $80, and $18 for eight babysitting jobs. how much did she
1 answer:
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Answer:
a) 81.5%
b) 95%
c) 75%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 266 days
Standard Deviation, σ = 15 days
We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.
Formula:
a) P(between 236 and 281 days)
b) a) P(last between 236 and 296)
c) If the data is not normally distributed.
Then, according to Chebyshev's theorem, at least data lies within k standard deviation of mean.
For k = 2
Atleast 75% of data lies within two standard deviation for a non normal data.
Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.
Answer:
A) Between 388 and 450
B) Between 357 and 481
C) Between 326 and 512
D) 99.995%
E) 72.907%
Step-by-step explanation:
We are given;
Mean; x¯ = 419 minutes
Standard deviation; σ = 31 minutes
A) From the empirical rule, 68% falls within 1 standard deviation.
Thus, it will be between;
419 - 1(31) and 419 + 1(31)
Thus gives;
Between 388 and 450
B) From the empirical rule, 95% falls within 2 standard deviation.
Thus, it will be between;
419 - 2(31) and 419 + 2(31)
Thus gives;
Between 357 and 481
C) From the empirical rule, 99.7% falls within 2 standard deviations.
Thus, it will be between;
419 - 3(31) and 419 + 3(31)
Thus;
Between 326 and 512
D)between 359 and 479 minutes, we will use z-s roe formula;
z = (479 - 359)/31
z = 3.871
From the z-table attached, we see that the probability is 0.99995 and as such, the percentage is 99.995%
E) worker spent 400 minutes on the job.
Thus;
z = (419 - 400)/31
z = 0.613
From z-tables, p = 0.72907
In percentage, we have; 72.907%
