Mathematics, 07.04.2020 20:13 mv603177

3. Suppose the local Best Buy store averages 301 customers every day entering the facility with a standard deviation of 80 customers. A random sample of 50 business days was selected. What is the probability that the average number of customers in the sample is between 290 and 310?

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Answer from: tyrique86

The probability that the average number of customers in the sample is between 290 and 310 is 0.61922.

Step-by-step explanation:

We are given that the local Best Buy store averages 301 customers every day entering the facility with a standard deviation of 80 customers.

A random sample of 50 business days was selected.

Let $\bar X$ = sample average number of customers

The z-score probability distribution for sample average is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population average = 301 customers

$\sigma$ = population standard deviation = 80 customers

n = sample of business days = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

(a) Probability the average number of customers in the sample is between 290 and 310 is given by = P(290 < $\bar X$ < 310) = P( $\bar X$ < 310) - P( $\bar X$ $\leq$ 290)

P( $\bar X$ < 310) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{310-301}{\frac{80}{\sqrt{50} } }$ ) = P(Z < 0.79) = 0.78524

P( $\bar X$ $\leq$ 290) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{290-301}{\frac{80}{\sqrt{50} } }$ ) = P(Z $\leq$ -0.97) = 1 - P(Z < 0.97)

= 1 - 0.83398 = 0.16602

{Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.79 and x = 0.97 in the z table which has an area of 0.78524 and 0.83398 respectively.}

Therefore, P(290 < $\bar X$ < 310) = 0.78524 - 0.16602 = 0.61922

Hence, the probability that the average number of customers in the sample is between 290 and 310 is 0.61922.

Answer from: Quest

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.

Answer from: Quest

You would need a laid out graph for this but could you provide a picture of just a graph? i can give stepssolve for y.subtract 4xdivide by -1 (opposites)graph slope intercept formcheck each point to see if it’s on the graph

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