Answer: Second option is correct.
Step-by-step explanation:
Since we have given that
where, N represents the number of bacteria after 'h' hours.
So, we have given that there are 450 bacteria present.
N = 450
According to question,
Hence, Second option is correct.
Answer:
(a) Null Hypothesis, :
41
Alternate Hypothesis, :
> 41
(b) The value of z test statistics is 1.08.
(c) We conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.
Step-by-step explanation:
We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.
It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.
Let = <u><em>mean sunspot activity during the Spanish colonial period.</em></u>
(a) Null Hypothesis, :
41 {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}
Alternate Hypothesis, :
> 41 {means that the mean sunspot activity during the Spanish colonial period was higher than 41}
The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean = 47
σ = population standard deviation = 35
n = sample of periods from Spanish colonial times = 40
So, <em><u>the test statistics</u></em> =
= 1.08
(b) The value of z test statistics is 1.08.
(c) <u>Now, the P-value of the test statistics is given by;</u>
P-value = P(Z > 1.08) = 1 - P(Z < 1.08)
= 1 - 0.8599 = <u>0.1401</u>
Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.
Therefore, we conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.
Answer:
2.28% of the class has a lower heart rate
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation(which is the square root of the variance)
, 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:
What percent of the class has a lower heart rate
Lower than 72, which is the pvalue of Z when X = 72.
has a pvalue of 0.0228
2.28% of the class has a lower heart rate