Achievement test scores from all high-school seniors in a certain state have a mean and variance of 60 and 64, respectively. A specific high school class of n=100 students had a mean score of 58. Is there evidence to suggest that this high school is substandard? (Consider calculating the probability that the sample mean is at most 58 when n=100)

0.62% that the sample mean is at most 58. Any probability lower than 5% is considered unusually low, which means that there is evidence to suggest that this high school is substandard.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

Also important to remember that the standard deviation is the square root of the variance. So

A specific high school class of n=100 students had a mean score of 58. Is there evidence to suggest that this high school is substandard? (Consider calculating the probability that the sample mean is at most 58 when n=100)

This probability is the pvalue of Z when X = 58. So

By the Central Limit Theorem

has a pvalue of 0.0062.

So there is a 0.62% that the sample mean is at most 58. Any probability lower than 5% is considered unusually low, which means that there is evidence to suggest that this high school is substandard.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

there are no choices to this, can you give us some choices so that i can answer it

step-by-step explanation: