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The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 73 minutes and a standard devi

Vesnalui [34]

Answer:

(a) $X\sim N(\mu = 73, \sigma = 16)$

(b) 0.7910

(c) 0.0401

(d) 0.6464

Step-by-step explanation:

Let X = amount of time that people spend at Grover Hot Springs.

The random variable X is normally distributed with a mean of 73 minutes and a standard deviation of 16 minutes.

(a)

The distribution of the random variable X is:

$X\sim N(\mu = 73, \sigma = 16)$

(b)

Compute the probability that a randomly selected person at the hot springs stays longer than 60 minutes as follows:

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-73}{16})\\=P(Z>-0.8125)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays longer than an hour is 0.7910.

(c)

Compute the probability that a randomly selected person at the hot springs stays less than 45 minutes as follows:

$P(X$

*Use a z-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays less than 45 minutes is 0.0401.

(d)

Compute the probability that a randomly person spends between 60 and 90 minutes at the hot springs as follows:

$P(60$

*Use a z-table for the probability.

Thus, the probability that a randomly person spends between 60 and 90 minutes at the hot springs is 0.6464

6 0

3 years ago

What is the simplest radical form of the expression? (x4y7)23

LUCKY_DIMON [66]

Answer:

23 x^ 4 y^ 7

Step-by-step explanation:

4 0

3 years ago

If DF = 9x-39, find EF.

otez555 [7]

$DF=DE+EF\\\\DF=9x-39\\DE=47\\EF=3x+10\\\\\text{substitute}\\\\9x-39=47+3x+10\\9x-39=3x+57\ \ \ |-3x\\6x-39=57\ \ \ \ |+39\\6x=96\ \ \ \ |:6\\x=16\\\\\text{put the value of x to the expression}\ 3x+10\\\\EF=3(16)+10=48+10=58$

Answer: EF = 58

7 0

3 years ago

Is algebra.

Nostrana [21]

Answer:

1) b 2) b

Step-by-step explanation:

1) Both expressions have (x+6). Rearrange them and you'll have one expression as (x+6) and the other as (5ab-4).

2) (4b - 7x)(a + b) factors to be 4ba + 8b - 7ax - 14x, which can be rearranged to 4ab - 7ax + 8b - 14x

3 0

3 years ago

Read 2 more answers

Researchers studying the acquisition of pronunciation often compare measurements made on the recorded speech of adults and child

dolphi86 [110]

Answer:

(A) The additional information that is needed to confirm about the conditions for this test have been met is ‘Population is approximately normal.

(B) The test statistic = 1.9965

(C1) P-value = 0.0556

(C2) There is no significant difference between mean VOT for children and adults.

D1) It is possible to make type II error.

(D2) There should be a difference in the VOT of adults and children.

Step-by-step explanation:

Let na be the number of adults = 20

xa mean VOT for adults = 88.17

sa standard deviatiation of VOT for adults = 24.74

Let nc be the number of children = 10

xc mean VOT for children =60.67

sc standard deviatiation of VOT for children = 39.89

(A) From the information the population variances are unknown and the two sample are assumed to be independent and the sample the sample size are smaller that is (n<30).

This indicates that the additional information that is required for the conditions of the test to be satisfied is 'distribution of the population'. the addition assumption to be made is, that the population distribution is normal.

(b) Calculating the test statistics using the formula;

t = (xa -xc)/SE - d

where SE = standard deviation , d= hypothesized difference = 0

But SE = √sa²/na +sc²/nc

= √24.74 ²/20 + 39.89²/10

= √189.72559

= 13.774

Substituting into test statistics equation, we have

t = (xa -xc)/SE - d

= (88.17 - 60.67/13.774

= 1.9965

Therefore the test statistic is 1.9965

(c) Calculating the p-value, we have;

Degree of freedom = na + nc -2

= 10+ 20 -2

= 28

The p-value for t=1.9965 at 28 degrees of freedom and 0.05 level of significance is 0.0556.

The p-value 0.0556 is greater than given level of significance 0.05 hence we fail to reject the null hypothesis and conclude that there is no significant difference between mean VOT for children and adults.

(D1) From the information in part (C) the null hypothesis is not rejected.

Since the null hypothesis is not rejected, there might be a chance that not rejecting null hypothesis would be wrong. In this type of situation the error that can occur would be type II error.

(D2) The type of error describe in the context of this study is obtained by the concept of the type II error which tells that the null hypothesis is not rejected when it is actually false.

7 0

3 years ago