Steve sold 36 fruit baskets for a school fundraiser. Evie sold 25% of the number of baskets that Steve sold. How many fruit bask
1 answer:
Answer:
9 fruit baskets.
Step-by-step explanation:
Given:
Steve sold 36 fruit baskets for a school fundraiser.
Evie sold 25% of the number of baskets that Steve sold.
<u>Question asked:</u>
How many fruit baskets did Evie sell?
Solution:
<u>As given that Evie sold 25% of the number of baskets that Steve sold.</u>
Number of baskets sold by Evie = 25% of 36
Thus, 9 fruit baskets sold by Evie.
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Answer:
(-0.1059 ; - 0.0337)
Step-by-step explanation:
The data table is attached in the picture below:
These is a matched pair design ; which requires taking the difference of the two values for each sample :
The mean and standard deviation of the difference will be used to construct the confidence interval :
The mean of difference, dbar = Σx/n = - 0.0698
The standard deviation of difference, Sd ;
Sd = [√Σ(d - dbar)²/(n-1)] = 0.1054
n = sample size = 25
The confidence interval :
dbar ± [TCritical * Sd/√n]
Tcritical at 90% ; df = n -1 = 25 -1
Tcritical(90% , 24) = 1.1711
C.I = - 0.0698 ± (1.711 * 0.1054/√25)
C.I = - 0.0698 ± 0.0361
C.I = (-0.1059 ; - 0.0337)
Answer:
a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.
c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that .
(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?
We are working with a sample mean of 37 jets. So we have that:
Total time of 320 minutes for 37 jets, so
This probability is the pvalue of Z when . So
has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?
Total time of 275 minutes for 37 jets, so
This probability is subtracted by the pvalue of Z when
has a pvalue of 0.0548.
There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.
(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?
Total time of 320 minutes for 37 jets, so
Total time of 275 minutes for 37 jets, so
This probability is the pvalue of Z when subtracted by the pvalue of Z when
.
So:
From a), we have that for , we have
, that has a pvalue of 0.7422.
From b), we have that for , we have
, that has a pvalue of 0.0548.
So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.