probability help needed

If he rides 60% of the time, the probability that he rides five days in a row is 60% to the fifth power.

But it only specifies that he rides 60% of the time, it never states that the proability he rides for one day is 60%?

He only goes to school once each day. No problem assuming that the probability he rides on any given day is 60%.

Thanks
Got a other question
What is the minimum number of times a fair coin must be tossed so that the probability of obtaining a tail on each toss is less than 0.0005
How would the above question be worked out?

The probability of obtaining n tails in a row is 2^(-n), so if you want to solve for n, you have the equation:

.0005 = 2^(-n)
log .0005 = -n log 2
-3.3010 = -n (.3010)
n = 10.96

So 11 tosses.

Thanks again
How would you find the median and mean of a hybrid function?

How do you tell if two events are mutually exclusive?

bump

This is in a significantly different category than your previous questions. Typically if you are given a function over a fixed domain, the mean is the integral of that function divided by size of the domain. Do you have an example?

Typically you are told or it is obvious from the question, but from a statistical point of view, you would try to correlate event two to event one and if there is no correlation at whatever confidence level you are interested in, you could say they are mutually exclusive.

The weight of a packaged orange cake is normally distributed with a mean of 500 g. If the packaged orange cake is more than 5 grams underweight, it is not acceptable. It has been found that 4% of the cakes are unacceptable. The standard deviation is therefore:

A 5.000

B 2.856

C 2.500

D 3.856

E None of the above

I'm really confused with the above question
I used the normal distribution on my case and none of the answers worked so I went with option E, none of the above
But the answer said it was B, 2.856
The answer converted this question to a Z score. Why is the above question converted to a Z score? I've seen similar questions to the one above and they didn't convert to Z score.
Any help would be appreciated
Thanks ;D

Any time a questions starts by saying you have a normal distribution, you are going to compute z scores. Z scores allow you to convert means and standard deviations to percentages which is exactly what this questions is asking.

In this question, you have a 4% failure rate. That is a z score of -1.75. (There are many z score calculators on the Internet if you don't have a table handy.) So five grams underweight is 1.75 standard deviations, so one standard deviation is 5/1.75 = 2.86 which is B.

Thanks
A supermarket claims that 90% of the strawberries it sells are regular. One of the customers found that there are 47 regular strawberries in his sample of 51 strawberries
Would the mean be 0.9 or 47/51
(Could someone please give an explanation of the reason as well)

Find the probability that atleast 85% of the strawberries are regular. 85% of 51 is 43.35, rounded up to 44
so n = 51, but is p = 90% or 47/51?

A new TV lottery game consists of the random selection, with replacement, of 3 marbles from a barrel. The 3
marbles are drawn from a group of 8 identical marbles except for colour. 3 of the marbles are blue and 5 are
yellow.
If X represents the number of blue marbles in the sample of 3, complete the probability distribution for X.
The above is a binomial distribution, would n be 3 or 8?
(please give reasong as well)
Thanks

A supermarket claims that 90% of all the strawberries it sells are regular.

One of the customers collects a random sample of 51 strawberries after buying two large packets and found that there are 47 regular strawberries in the sample



a. a. Find the exact expected value and standard deviation, of the sampling distribution of the proportion of regular strawberries in his sample

Is the expected value 0.9 or 47/51? Please explain why



When calculating the standard deviation would we use 0.9 or 47/51? Please explain why



Find the probability that 85% of the strawberries in his sample are regular

85% of 51 = 43.35, which rounds up to 44, but do we use 0.9 or 47/51? Please explain why



Use the normal distribution with the expected value and standard deviation from part a. to find the approximate probability that at least 80% of the strawberries in his sample are regular. Give your answer correct to three decimal places

Since it says to use the normal distribution is the expected value 0.9 * 51 = 45.9 and the standard deviation the square root (45.9*0.1)?



Find the z value that gives a 98% confidence interval. How would you convert confidence intervals to z values?

Find the 98% confidence interval for the proportion of reagular strawberries sold by the supermarket

Do we use 0.9 or 47/51? Please explain why



As you can probably tell, I’m confused with this sampling topic in probability and would appreciate any help. Thanks