A medium pizza costs $8.60. Darcy has $54 to spend on pizzas. How many medium pizzas can she afford?
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Answer:
We conclude that the mean contents of cola bottles is more than or equal to the advertised 300 ml.
Step-by-step explanation:
We are given that Bottles of a popular cola drink are supposed to contain 300 ml of cola. The distribution of the contents is normal with standard deviation of 3 ml.
A student who suspects that the bottler is under-filling measures the contents of six bottles. The results are: 299.4, 297.7, 301.0, 298.9, 300.2, 297.0
<u><em>Let </em></u><u><em> = mean contents of cola bottles.</em></u>
SO, Null Hypothesis, :
300 ml {means that the mean contents of cola bottles is more than or equal to the advertised 300 ml}
Alternate Hypothesis, :
< 300 ml {means that the mean contents of cola bottles is less than the advertised 300 ml}
The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about the population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean contents of cola bottle =
= 299.03 ml
= population standard deviation = 3 ml
n = sample of bottles = 6
So, <em><u>test statistics</u></em> =
= -0.792
<em>Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -0.792 > -1.6449, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>
Therefore, we conclude that the mean contents of cola bottles is more than or equal to the advertised 300 ml.
The probability that the first two votes drawn are both for candidate a is given by:
3C2/5C2 = 3/10
Having drawn two votes for candidate a on the first two draws, there are 2 votes for candidate b and one vote for candidate a remaining. The probability that a vote for candidate b will be drawn on the third draw is:
2/3.
After the first three draws, there reains one vote for candidate a and one vote for candidate b. The probability that a vote for candidate a will be drawn on the fourth draw is:
1/2.
The probability of the ordering aabab is therefore given by:
The answer is: 0.1.
3C2/5C2 = 3/10
Having drawn two votes for candidate a on the first two draws, there are 2 votes for candidate b and one vote for candidate a remaining. The probability that a vote for candidate b will be drawn on the third draw is:
2/3.
After the first three draws, there reains one vote for candidate a and one vote for candidate b. The probability that a vote for candidate a will be drawn on the fourth draw is:
1/2.
The probability of the ordering aabab is therefore given by:
The answer is: 0.1.