Identify terms and factors: - 4x + 2x2y
2 answers:
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Answer:
The percentage of people should be seen by the doctor between 13 and
17 minutes is 68% ⇒ 2nd term
Step-by-step explanation:
* Lets explain how to solve the problem
- Wait times at a doctor's office are typically 15 minutes, with a standard
deviation of 2 minutes
- We want to find the percentage of people should be seen by the
doctor between 13 and 17 minutes
* To find the percentage we will find z-score
∵ The rule the z-score is z = (x - μ)/σ , where
# x is the score
# μ is the mean
# σ is the standard deviation
∵ The mean is 15 minutes and standard deviation is 2 minutes
∴ μ = 15 , σ = 2
∵ The people should be seen by the doctor between 13 and
17 minutes
∵ x = 13 and 17
∴ z =
∴ z =
- Lets use the standard normal distribution table
∵ P(z > -1) = 0.15866
∵ P(z < 1) = 0.84134
∴ P(-1 < z < 1) = 0.84134 - 0.15866 = 0.68268 ≅ 0.68
∵ P(13 < x < 17) = P(-1 < z < 1)
∴ P(13 < x < 17) = 0.68 × 100% = 68%
* The percentage of people should be seen by the doctor between
13 and 17 minutes is 68%
Answer:
(a) The median is 1478
(b) Null hypothesis H₀ : μ₁ = μ₂
Alternative hypothesis Hₐ : μ₁ > μ₂
(c) The test statistic is 6.678155
(d) The p value is 1.2×10⁻¹¹
(e) We reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high
Step-by-step explanation:
(a) Here we have the data as follows;
1623 1586 1574 1554 1544 1531 1519 1513 1491 1484 1478 1471 1458 1456 1458 1453 1438 1428 1405 1395 1377
The median is = 1478
Therefore we have above the median
1623 1586 1574 1554 1544 1531 1519 1513 1491 1484
The mean, = 1541.9
Standard deviation, σ₁ = 41.38224257
n₁ = 10
Below the median
1471 1458 1456 1458 1453 1438 1428 1405 1395 1377
The mean, = 1433.9
Standard deviation, σ₂ = 30.04812806
n₂ = 10
(b) Null hypothesis H₀ : μ₁ = μ₂
Alternative hypothesis Hₐ : μ₁ > μ₂
(c) The formula for t test is given by;
df = 10 - 1 = 9, α = 0.05
Therefore, the test statistic = 6.678155
(d) The p value from statistical relations is Probability p = 1.2×10⁻¹¹
Critical z at 5% confidence level = 1.645
Since P << 0.05, e reject the
(e) Therefore, we reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high.