To check whether there is a reason to believe that the average height has been changed, we can check one right-tailed test and establish null and alternate hypotheses.
H₀ (null hypothesis): μ = 162.5
H₁ (alternate hypothesis): μ > 162.5
Since we have two samples to compare, we can use the formula below to compute for the z-score.

where Z is the z-score, Χ is the new mean of the sample, μ is the expected average value, δ is the standard deviation, and n is the sample size. Given the values we have,


Assuming that the significance level, α, is 0.05. Now that we have a z-score of 2.77, we can use the z-table to find its p-value. Thus, we have P(Z > 2.77) = .0028. Thus since our p-value is less than α, H₀ is rejected. That means that the average height of the female freshman students has changed.
Answer:
x = 20
Step-by-step explanation:
Simplifying
1250 + 27.50x = 1400 + 20x
Solving
1250 + 27.50x = 1400 + 20x
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-20x' to each side of the equation.
1250 + 27.50x + -20x = 1400 + 20x + -20x
Combine like terms: 27.50x + -20x = 7.5x
1250 + 7.5x = 1400 + 20x + -20x
Combine like terms: 20x + -20x = 0
1250 + 7.5x = 1400 + 0
1250 + 7.5x = 1400
Add '-1250' to each side of the equation.
1250 + -1250 + 7.5x = 1400 + -1250
Combine like terms: 1250 + -1250 = 0
0 + 7.5x = 1400 + -1250
7.5x = 1400 + -1250
Combine like terms: 1400 + -1250 = 150
7.5x = 150
Divide each side by '7.5'.
x = 20
Simplifying
x = 20
Divide by 10 to find how many tens he needs.
1200/10 = 120
Answer: He needs 120 tens
Answer:
a.
b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349
Step-by-step explanation:
a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.
b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.
c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842
d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166
e. The z-score related to 6.4 kg is
and the z-score related to 7 kg is
, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194
f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349