What are the answers fast someone please
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Answer:
Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.
Step-by-step explanation:
We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.
A sample of 36 students was selected.
<em>Let </em><em> = sample average time spent studying</em>
The z-score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean hours spent studying = 25 hours
= standard deviation = 15 hours
n = sample of students = 36
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < < 30 hours)
P(29 hours < < 30 hours) = P(
< 30 hours) - P(
29 hours)
P( < 30 hours) = P(
<
) = P(Z < 2) = 0.97725
P(
29 hours) = P(
) = P(Z
1.60) = 0.94520
<em>So, in the z table the P(Z </em><em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.</em>
Therefore, P(29 hours < < 30 hours) = 0.97725 - 0.94520 = 0.0321
Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.
Answer:
There is no sufficient evidence to reject the company's claim at the significance level of 0.05
Step-by-step explanation:
Let be the true mean weight per apple the company ship. We want to test the next hypothesis
vs
(two-tailed test).
Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by
which is normally distributed. The observed value is
. The rejection region for
is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.
Answer:
(- 7, - 2 ) , (9.5, 0.75 )
Step-by-step explanation:
Given the equations
x - 6y = 5 ( add 6y to both sides )
x = 5 + 6y → (1)
xy - 2y² = 6 → (2)
Substitute x = 5 + 6y into (2)
(5 + 6y)y - 2y² = 6 ← distribute and rearrange left side
5y + 6y² - 2y² = 6
4y² + 5y = 6 ( subtract 6 from both sides )
4y² + 5y - 6 = 0 ← in standard form
(y + 2)(4y - 3) = 0 ← in factored form
Equate each factor to zero and solve for y
y + 2 = 0 ⇒ y = - 2
4y - 3 = 0 ⇒ 4y = 3 ⇒ y = = 0.75
Substitute these values into (1) for corresponding values of x
y = - 2 : x = 5 + 6(- 2) = 5 - 12 = - 7 ⇒ (- 7, - 2 )
y = 0.75 : x = 5 + 6(0.75) = 5 + 4.5 = 9.5 ⇒ (9.5, 0.75 )