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Vaselesa [24]
1 year ago
8

The sales data for January and February of a frozen yogurt shop are approximately normal. The mean daily sales for January was $

300 with a standard deviation of $20. On the 15th of January, the shop sold $310 of yogurt. The mean daily sales for February was $320 with a standard deviation of $50. On the 15th of February, the shop sold $340 of yogurt. Which month had a higher z-score for sales on the 15th, and what is the value of that z-score
Mathematics
1 answer:
VashaNatasha [74]1 year ago
3 0

Answer:

January had a higher z-score for sales on the 15th, and the value of that z-score was of 0.5.

Step-by-step explanation:

z-score:

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

January:

The mean daily sales for January was $300 with a standard deviation of $20. On the 15th of January, the shop sold $310 of yogurt. This means, respectively, that \mu = 300, \sigma = 20, X = 310. So

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

Z = \frac{310 - 300}{20}

Z = 0.5

February:

The mean daily sales for February was $320 with a standard deviation of $50. On the 15th of February, the shop sold $340 of yogurt. This means, respectively, that \mu = 320, \sigma = 50, X = 340. So

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

Z = \frac{340 - 320}{50}

Z = 0.4

January had a higher z-score for sales on the 15th, and the value of that z-score was of 0.5.

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Answer:

As per the given statement:

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From the given figure:

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5 0
1 year ago
The heights of 82 roller coasters have a mean of 285.2 feet and a population standard deviation of 59.3 feet. Find the standardi
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Answer:

0.7673

Step-by-step explanation:

We have the following:

The null and alternative hypothesis is,

H0: m = 290

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is m the mean, sd the standard deviation and n the population size

Now we calculate the value of z like this:

z = (x - m) / sd / (n ^ (1/2))

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7 0
1 year ago
Ramina found the length of two pieces of ribbon to be 47.6 inches and 39.75 inches. Which is an effective estimation strategy fo
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Answer:

<h3>Add 47.6 and 39.75, then round the answer</h3>

Step-by-step explanation:

If Ramina found the length of two pieces of ribbon to be 47.6 inches and 39.75 inches, the effective strategy of finding the sum of the two lengths is to:

1) First is to add the two values together

47.6 + 39.75

= (47+0.6)+(39+0.75)

= (47+39)+(0.6+0.75)

= 86 + 1.35

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2) Round up the answer to nearest whole number.

87.35 ≈ 87 (note that we couldn't round up to 88 because the values after the decimal point wasn't up to 5)

Option C is correct

7 0
1 year ago
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The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,
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Answer:

A.the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

\mathtt{H_o: \mu = 100}

\mathtt{H_1: \mu \neq 100}

A. If the acceptance region is defined as 98.5 <  \overline x >  101.5 , find the type I error probability \alpha .

Assuming the critical region lies within \overline x < 98.5 or \overline x > 101.5, for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is \mu = 100

∴

\mathtt{\alpha = P( type  \ 1  \ error ) = P( reject \  H_o)}

\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5  )}

when  \mu = 100

\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}}  > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }

\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z  > \dfrac{1.5}{\dfrac{2}{3}}) }

\mathtt{\alpha = P ( Z  2.25) }

\mathtt{\alpha = P ( Z

From the standard normal distribution tables

\mathtt{\alpha = 0.0122+( 1-  0.9878) })

\mathtt{\alpha = 0.0122+( 0.0122) })

\mathbf{\alpha = 0.0244 }

Thus, the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis \mathtt{H_o}

Thus;

β = P( type II error) - P( fail to reject \mathtt{H_o} )

∴

\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

Given that \mu = 103

\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }

\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

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Given that \mu = 105

\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }

\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

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