Please help me on this question.

Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words pe

VikaD [51]

Answer:

36.04% probability that at most two of them would read at less than 850 words per minute

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the binomial probability distribution.

Normal probability distribution

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

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Percentage of students who read less than 850 words per minute.

Pvalue of Z when X = 850. The mean is $\mu = 950$ and the standard deviation is $\sigma = 200$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{850 - 950}{200}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085.

30.85 of students read less than 850 words per minute.

If 10 students are selected at random, what is the probability that at most two of them would read at less than 850 words per minute

This is $P(X \leq 2)$ when $n = 10, p = 0.3085$ . So

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.3085)^{0}.(0.6915)^{10} = 0.0250$

$P(X = 1) = C_{10,1}.(0.3085)^{1}.(0.6915)^{9} = 0.1115$

$P(X = 2) = C_{10,2}.(0.3085)^{2}.(0.6915)^{8} = 0.2239$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0250 + 0.1115 + 0.2239 = 0.3604$

36.04% probability that at most two of them would read at less than 850 words per minute

8 0

3 years ago

Given: LMNP is a square A, B midpoints of LM and PN Q ∈ AB m∠LPQ = 45° Find: m∠PQN

lisov135 [29]

Since A and B are the midpoints of ML and NP, we can say that AB is parallel to MN and LP. In order to find ∠PQN, we can work with the triangles PQB and NQB. According to SAS (Side-Angle-Side) principle, these triangles are congruent. BQ is a common side for these triangles and NB=BP and the angle between those sides is 90°, i.e, ∠NBQ=∠PBQ=90°. After finding that these triangles are equal, we can say that ∠BNQ is 45°. From here, we easily find <span>∠PQN. It is 180 - (</span>∠QNP + ∠NPQ) = 180 - 90 = 90°