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<h3>HCF of 4 and 9 is 1</h3>
- HCF of 4 and 9 is the largest possible number that divides 4 and 9 exactly without any remainder. The factors of 4 and 9 are 1, 2, 4 and 1, 3, 9 respectively.
So , the HCF of 4 and 9 is 1
Answer:
Since we can't assume that the distribution of X is the normal then we need to apply the central limit theorem in order to approximate the with a normal distribution. And we need to check if n>30 since we need a sample size large as possible to assume this.
Based on this rule we can conclude:
a. n = 14 b. n = 19 c. n = 45 d. n = 55 e. n = 110 f. n = 440
Only for c. n = 45 d. n = 55 e. n = 110 f. n = 440 we can ensure that we can apply the normal approximation for the sample mean
for n=14 or n =19 since the sample size is <30 we don't have enough evidence to conclude that the sample mean is normally distributed
Step-by-step explanation:
For this case we know that for a random variable X we have the following parameters given:
Since we can't assume that the distribution of X is the normal then we need to apply the central limit theorem in order to approximate the with a normal distribution. And we need to check if n>30 since we need a sample size large as possible to assume this.
Based on this rule we can conclude:
a. n = 14 b. n = 19 c. n = 45 d. n = 55 e. n = 110 f. n = 440
Only for c. n = 45 d. n = 55 e. n = 110 f. n = 440 we can ensure that we can apply the normal approximation for the sample mean
for n=14 or n =19 since the sample size is <30 we don't have enough evidence to conclude that the sample mean is normally distributed