laura has 160 pounds of green beans at her vegetable stand. she sells 10 pounds of green beans every hour, what is the percent d
1 answer:
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Answer:
a)0.6192
b)0.7422
c)0.8904
d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.
Step-by-step explanation:
Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then
z(p)= where
- Me is the margin of error from the mean
- s is the standard deviation of the population
- N is the sample size
a.
z(p)= ≈ 0.8764
by looking z-table corresponding p value is 1-0.3808=0.6192
b.
z(p)= ≈ 1.1314
by looking z-table corresponding p value is 1-0.2578=0.7422
c.
z(p)= ≈ 1.6
by looking z-table corresponding p value is 1-0.1096=0.8904
d.
Minimum required sample size for 0.95 probability is
N≥ where
- N is the sample size
- z is the corresponding z-score in 95% probability (1.96)
- s is the standard deviation (50)
- ME is the margin of error (8)
then N≥ ≈150.6
Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.