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Using the normal approximation to the binomial distribution, it is found that:
a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.
b) Since Z < 2, 717 peas with red flowers is not significantly high.
c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.
For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.
Binomial distribution:
Probability of x successes on n trials, with p probability.
Normal distribution:
In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
- If Z > 2, the result is considered <u>significantly high</u>.
If and
, the binomial distribution can be approximated to the normal with:
In this problem:
- 943 peas, thus,
- 3/4 probability of being red, thus
.
Applying the approximation:
Item a:
Using continuity correction, this probability is , which is <u>1 subtracted by the p-value of Z when X = 716.5</u>.
Then:
has a p-value of 0.758.
1 - 0.758 = 0.242
0.242 = 24.2% probability of getting 717 or more peas with red flowers.
Item b:
Since Z < 2, 717 peas with red flowers is not significantly high.
Item c:
Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.
A similar problem is given at brainly.com/question/25212369