Please help me out :)
1 answer:
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Step-by-step explanation:
Assuming the data is as shown, restaurant X has a mean service time of 180.56, with a standard deviation of 62.6.
The standard error is SE = s/√n = 62.6/√50 = 8.85.
At 95% confidence, the critical value is z = 1.960.
Therefore, the confidence interval is:
180.56 ± 1.960 × 8.85
180.56 ± 17.35
(163, 198)
Restaurant Y has a mean service time of 152.96, with a standard deviation of 49.2.
The standard error is SE = s/√n = 49.2/√50 = 6.96.
At 95% confidence, the critical value is z = 1.960.
Therefore, the confidence interval is:
152.96 ± 1.960 × 6.96
152.96 ± 13.64
(139, 167)
Answer: It is such a small percentage of the population that would be considered to have fever at that level.
So, it is not appropriate stringent standard.
Step-by-step explanation:
Since we have given that
Mean = 98.20°F
Standard deviation = 0.62°F
If Bellevue Hospital in New York City uses 100.6°F as the lowest temperature considered to be a fever.
So,
So, using the normal distribution, we first find the value of z.
Since z = 3.87
So, p = 0.0001 =0.1%
So,It is such a small percentage of the population that would be considered to have fever at that level.
So, it is not appropriate stringent standard.