Question

One year consumers spent an average of ​$24 on a meal at a restaurant. Assume that the amount spent on a restaurant meal is normally distributed and that the standard deviation is ​$4.

Between what two​ values, symmetrically distributed around the​ mean, will the middle 95​% of the amounts of cash spent​ fall?.

Answer 1

The middle 95% of the amounts of cash spent on a restaurant meal fall between $16.16 and $31.84, calculated using the normal distribution with a mean of $24 and a standard deviation of $4.

Step-by-step explanation:

To find the two values symmetrically distributed around the mean where the middle 95% of the amounts spent on a restaurant meal fall, we will use the properties of the normal distribution. The average amount spent is $24 with a standard deviation of $4. For a normal distribution, the middle 95% is often represented by the range from the mean minus 1.96 standard deviations to the mean plus 1.96 standard deviations, because 1.96 standard deviations from the mean approximately captures the central 95% of data in a normal distribution.

Step-by-step calculation:

1. Calculate the lower boundary: Mean - (1.96 * Standard Deviation) = $24 - (1.96 * $4).
2. Calculate the upper boundary: Mean + (1.96 * Standard Deviation) = $24 + (1.96 * $4).

Performing the calculations:

• Lower boundary = $24 - $7.84 = $16.16.
• Upper boundary = $24 + $7.84 = $31.84.

Therefore, the middle 95% of the amounts of cash spent fall between $16.16 and $31.84.